Limits (Real Functions)#

Real One-Sided Limits#

Definition: Left-Sided Limit

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function, let \(c \in \mathbb{R}\) be an accumulation point of \(\mathcal{D} \cap (-\infty; c)\) and let \(L \in \mathbb{R}\).

We say that \(L\) is the left-sided limit of \(f\) at \(c\) if for every \(\varepsilon \gt 0\) there is some \(\delta \gt 0\) such that

\[|x-c| \lt \delta \implies |f(x)-L| \lt \varepsilon\]

for all \(x \in \mathcal{D}\) with \(x \lt c\).

Notation

\[\lim_{x\to c^-} f(x) = L \qquad \lim_{\begin{matrix} x \to c \\ x \lt c\end{matrix}} f(x) = L\]

Theorem: Left-Sided Limit via Sequences

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function, let \(c \in \mathbb{R}\) be an accumulation point of \(\mathcal{D} \cap (-\infty; c)\) and let \(L \in \mathbb{R}\).

The following statements are equivalent:

The left-sided limit of \(f\) at \(c\) is \(L\):

\[\lim_{x \to c^{-}} f(x) = L\]

For each infinite sequence \((x_n)_{n \in \mathcal{D}'}\) with \(x_n \lt c\) for all \(n \in \mathcal{D}'\) which converges to \(c\), the sequence \((f(x_n))_{n \in \mathcal{D}'}\) converges to \(L\):

\[\lim_{n \to \infty} x_n = c \implies \lim_{n \to \infty} f(x_n) = L\]

Proof

TODO

Definition: Right-Sided Limit

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function, let \(c \in \mathbb{R}\) be an accumulation point of \(\mathcal{D} \cap (c; +\infty)\) and let \(L \in \mathbb{R}\).

We say that \(L\) is the right-sided limit of \(f\) at \(c\) if for every \(\varepsilon \gt 0\) there is some \(\delta \gt 0\) such that

\[|x-c| \lt \delta \implies |f(x)-L| \lt \varepsilon\]

for all \(x \in \mathcal{D}\) with \(x \gt c\).

Notation

\[\lim_{x\to c^+} f(x) = L \qquad \lim_{\begin{matrix} x \to c \\ x \gt c\end{matrix}} f(x) = L\]

Theorem: Right-Sided Limit via Sequences

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function, let \(c \in \mathbb{R}\) be an accumulation point of \(\mathcal{D} \cap (c; +\infty)\) and let \(L \in \mathbb{R}\).

The following statements are equivalent:

The right-sided limit of \(f\) at \(c\) is \(L\):

\[\lim_{x \to c^{+}} f(x) = L\]

For each infinite sequence \((x_n)_{n \in \mathcal{D}'}\) with \(x_n \gt c\) for all \(n \in \mathcal{D}'\) which converges to \(c\), the sequence \((f(x_n))_{n \in \mathcal{D}'}\) converges to \(L\):

\[\lim_{n \to \infty} x_n = c \implies \lim_{n \to \infty} f(x_n) = L\]

Proof

TODO

Real Limits#

Definition: Limit of a Function (Cauchy)

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(c \in \mathbb{R}\) be an accumulation point of \(\mathcal{D}\).

We say that \(L \in \mathbb{R}\) is the limit of \(f\) at \(c\) if for each \(\varepsilon \gt 0\), there exists some \(\delta \gt 0\) such that for all \(x \in \mathcal{D}\), we have the following:

\[0 \lt |x-c| \lt \delta \implies |f(x)-L| \lt \varepsilon\]

Notation

\[\lim_{x\to c} f(x) = L\]

Definition: Limit at Positive Infinity

Let \(\mathcal{D}\) be unbounded above and let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

We say that \(L\) is the limit of \(f\) at positive infinity if for each \(\varepsilon \gt 0\) there is some \(A \in \mathbb{R}\) such that

\[|f(x) - L| \lt \varepsilon\]

for all \(x \in \mathcal{D}\) with \(x \ge A\).

Notation

\[\lim_{x\to \infty} f(x) = L\]

Example: \(\lim_{x \to +\infty} \frac{1}{x}\)

The limit of \(\frac{1}{x}\) at \(+\infty\) is \(0\):

\[\lim_{x \to +\infty} \frac{1}{x} = 0\]

Definition: Limit at Negative Infinity

Let \(\mathcal{D}\) be unbounded below and let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

We say that \(L\) is the limit of \(f\) at negative infinity if for each \(\varepsilon \gt 0\) there is some \(A \in \mathbb{R}\) such that

\[|f(x) - L| \lt \varepsilon\]

for all \(x \in \mathcal{D}\) with \(x \le A\).

Notation

\[\lim_{x\to -\infty} f(x) = L\]

Definition: Limit Existence

If \(\lim_{x \to c} f(x) = L \in \mathbb{R}\) for \(c \in \mathbb{R} \cup \{-\infty, +\infty\}\), then we say that the limit of \(f\) at \(c\) exists.

Theorem: Real Limit and One-Sided Real Limits

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(c \in \mathbb{R}\) be an accumulation point of both \(\mathcal{D} \cap (-\infty;c)\) and \(\mathcal{D} \cap (c; +\infty)\).

The two-sided limit of \(f\) at \(c\) exists if and only if both one-sided limits of \(f\) at \(c\) exist and are equal.

\[\lim_{x \to c} f(x) = L \in \mathbb{R} \iff \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L\]

Proof

TODO

Example: The Floor Function

The floor function is the function \(\lfloor \cdot\rfloor: \mathbb{R} \to \mathbb{R}\) which for each \(x\) gives the largest integer which is less than or equal to \(x\). For example,

\[\lfloor 0.5 \rfloor = 0 \qquad \lfloor 1.2 \rfloor = 1 \qquad \lfloor -3.3 \rfloor = -4 \qquad \lfloor 5 \rfloor = 5\]

For each integer \(m \in \mathbb{Z}\) we have

\[\lim_{x \to m^{-}} \lfloor x \rfloor = m - 1 \qquad \lim_{x \to m^{+}} \lfloor x \rfloor = m.\]

This means that \(\lim_{x \to m} \lfloor x \rfloor\) does not exist when \(m\) is an integer.

Theorem: Limits through Transformations

Let \(f\) and \(g\) be real functions and let \(c \in \mathbb{R}\).

If there exists some deleted neighborhood of \(c\) on which \(f\) and \(g\) are equal and the limit of \(g\) at \(c\) is \(L \in \mathbb{R}\), then

\[\lim_{x \to c} f(x) = \lim_{x \to c} g(x) = L.\]

Note

This theorem is extremely powerful because it allows us to find the limit of \(f\) by finding another function \(g\) (usually through algebraic manipulations) whose limit is easier to compute, so long as the two functions are equal around \(c\). We don't care about what happens at \(x = c\) or if \(f\) and \(g\) are even defined at \(c\).

Proof

Let \(\varepsilon \gt 0\) be any positive real number.

Since there is some deleted neighborhood of \(c\) on which \(f\) and \(g\) are equal, we know that there exists some \(\delta_1 \gt 0\) such that

\[0 \lt |x - c| \lt \delta_1 \implies f(x) = g(x).\]

Since \(\lim_{x \to c} g(x) = L\), we know that for our choice of \(\varepsilon\) there exists some \(\delta_2 \gt 0\) such that

\[0 \lt |x - c| \lt \delta_2 \implies |g(x) - L| \lt \varepsilon.\]

Set \(\delta = \min\{\delta_1, \delta_2\}\).

Since \(\delta \le \delta_1\), we have

\[0 \lt |x - c| \lt \delta \implies 0 \lt |x - c| \lt \delta\]

and, since \(0 \lt |x - c| \lt \delta_1 \implies f(x) = g(x)\), we have proved that

\[0 \lt |x - c| \lt \delta \implies f(x) = g(x).\]

Similarly, since \(\delta \le \delta_2\), we have

\[0 \lt |x - c| \lt \delta \implies 0 \lt |x - c| \lt \delta\]

and, since \(0 \lt |x - c| \lt \delta_2 \implies |g(x) - L| \lt \varepsilon\), we have proved that

\[0 \lt |x - c| \lt \delta \implies |g(x) - L| \lt \varepsilon.\]

Now we combine these facts. Consider \(|f(x) - L|\). For all \(x\) with \(0 \lt |x - c| \lt \delta\), we showed that \(f(x) = g(x)\). Therefore,

\[0 \lt |x - c| \lt \delta \implies |f(x) - L| = |g(x) - L|\]

Moreover, we showed that \(|g(x) - L| \lt \varepsilon\) for all \(x\) with \(0 \lt |x - c| \lt \delta\). Therefore,

\[0 \lt |x - c| \lt \delta \implies |f(x) - L| \lt \varepsilon.\]

Example

Consider the function \(f: \mathbb{R} \setminus \{1\} \to \mathbb{R}\) defined in the following way:

\[f(x) = \frac{x^3 - 1}{x-1}\]

Consider also the function \(g: \mathbb{R} \to \mathbb{R}\) defined in the following way:

\[g(x) = x^2 + x + 1\]

We want to determine the [limits](./Limits%20(Real%20Functions.md) of \(f\) and \(g\) at \(c = 1\). We see, for example, that \(f\) and \(g\) are equal on the deleted neighborhood \((0; 1)\cup(1;2)\) of \(c\), since

\[f(x) = \frac{x^3 - 1}{x - 1} = \frac{(x-1)(x^2+x+1)}{x-1} = x^2+x+1\]

for all \(x \ne 1\). We thus know that the [limits](./Limits%20(Real%20Functions.md) of \(f\) and \(g\) at \(c\), if they exist, must be equal:

\[\lim_{x \to c} f(x) = \lim_{x \to c} g(x)\]

Now, \(g\) is a real polynomial function and so we know that

\[\lim_{x \to c} g(x) = g(c) = 1^2 + 1 + 1 = 3.\]

This automatically means that

\[\lim_{x \to c} f(x) = 3.\]

Example

Consider the function \(f: \mathbb{R} \setminus \{0\} \to \mathbb{R}\) defined in the following way:

\[f(x) = \frac{\sqrt{x + 1} - 1}{x}\]

We want to determine its [limit](./Limits%20(Real%20Functions.md) at \(0\).

For \(x \ne 0\), we have

\[\begin{aligned} f(x) & = \frac{\sqrt{x + 1} - 1}{x} \\ &= \frac{(\sqrt{x + 1} - 1)(\sqrt{x + 1} + 1)}{x(\sqrt{x + 1} + 1)} \\ &= \frac{1 + x - 1}{x(\sqrt{x + 1} + 1)} \\ &= \frac{x}{x(\sqrt{x + 1} + 1)} \\ &= \frac{1}{\sqrt{x+1}+1} \qquad \forall x \in \mathbb{R} \setminus \{0\}\end{aligned}\]

Since \(\mathbb{R} \setminus \{0\}\) is a deleted neighborhood of \(0\), we know that

\[\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x} = \lim_{x \to 0} \frac{1}{\sqrt{x+1}+1}\]

The function \(\frac{1}{\sqrt{x+1}+1}\) is continuous at \(0\), i.e.

\[\lim_{x \to 0} \frac{1}{\sqrt{x+1}+1} = \frac{1}{\sqrt{1 + 1} + 1} = \frac{1}{2}.\]

Therefore,

\[\lim_{x \to 0} f(x) = \frac{1}{2}.\]

Theorem: Arithmetic with Real Limits

Let \(f\) and \(g\) be real functions and let \(c \in \mathbb{R} \cup \{-\infty, +\infty\}\).

If the [limits](./Limits%20(Real%20Functions.md) of \(f\) and \(g\) exist at \(c\), then

\[\begin{aligned} &\lim_{x\to c} \left(\alpha f(x) + \beta g(x)\right) = \alpha \lim_{x \to c} f(x) + \beta \lim_{x \to c} g(x) \qquad \forall \alpha, \beta \in \mathbb{R} \\ \\ &\lim_{x \to c} \left(f(x) g(x)\right) = \left(\lim_{x\to c} f(x)\right) \cdot \left(\lim_{x\to c} g(x)\right) \\ \\ &\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\displaystyle \lim_{x \to c} f(x)}{\displaystyle \lim_{x \to c} g(x)}, \qquad \text{if} \lim_{x \to c} g(x) \ne 0 \\ \\ & \lim_{x \to c} f(x)^{g(x)} = \lim_{x \to c} f(x)^{\displaystyle \lim_{x \to c} g(x)}, \qquad \text{if} \lim_{x \to c} f(x) \gt 0\end{aligned}\]

Proof

TODO

Warning

These do not apply to infinite limits.

Theorem: The Squeeze Theorem for Functions

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\), \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) and \(h: \mathcal{D}_h \subseteq \mathbb{R} \to \mathbb{R}\) be real functions and let \(c \in \mathbb{R}\).

If there exists some deleted neighborhood of \(c\) on which \(f(x) \le h(x) \le g(x)\) and the limits of \(f\) and \(g\) exist at \(c\) and are equal to the same \(L \in \mathbb{R}\), then

\[\lim_{x\to c} f(x) = \lim_{x \to c} h(x) = \lim_{x \to c} g(x) = L.\]

Proof

TODO

Example: \(\lim_{x \to 0} \frac{\sin x}{x} = 1\)

We want to find the [limit](./Limits%20(Real%20Functions.md) at \(0\) of the function \(f: \mathbb{R} \setminus \{0\}\to \mathbb{R}\) defined in the following way:

\[f(x) = \frac{\sin x}{x}\]

We examine the deleted neighborhood \(\mathcal{N} = \left(-\frac{\pi}{2}; 0\right)\cup\left(0; +\frac{\pi}{2}\right)\).

TODO

Example: \(\lim_{x \to 0} \frac{\mathrm{e}^x - 1}{x} = 1\)

We want to find the [limit](./Limits%20(Real%20Functions.md) at \(0\) of the function \(f: \mathbb{R} \setminus \{0\}\to \mathbb{R}\) defined in the following way:

\[f(x) = \frac{\mathrm{e}^x - 1}{x}\]

TODO

Example: \(\lim_{x \to 0} x \sin \frac{1}{x}\)

We want to find the [limit](./Limits%20(Real%20Functions.md) at \(0\)of the function \(f: \mathbb{R} \setminus \{0\}\to \mathbb{R}\) defined in the following way:

\[f(x) = x \sin \frac{1}{x}\]

Since the image of sin is \([-1; +1]\), we have

\[-|x| \le x \sin \frac{1}{x} \le +|x| \qquad \forall x \in \mathbb{R} \setminus \{0\}.\]

We know that \(\mathbb{R} \setminus \{0\}\) is a deleted neighborhood of \(0\) and, since \(\lim_{x \to 0} -|x| = \lim_{x \to 0} +|x| = 0\), we have

\[\lim_{x \to 0} x\sin\frac{1}{x} = 0.\]

Theorem: Limits Inequality

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be real functions.

If the limits of \(f\) and \(g\) exist at \(c \in \mathbb{R}\) and there exists some deleted neighborhood of \(c\) on which \(g(x) \le f(x)\) and, then

\[\lim_{x \to c} g(x) \le \lim_{x \to c} f(x).\]

If the limits of \(f\) and \(g\) exist at \(- \infty\) and there exists some \(A \in \mathbb{R}\) such that \(g(x) \le f(x)\) for all \(x \lt A\), then

\[\lim_{x \to -\infty} g(x) \le \lim_{x \to -\infty} f(x).\]

If the limits of \(f\) and \(g\) exist at \(+ \infty\) and there exists some \(B \in \mathbb{R}\) such that \(g(x) \le f(x)\) for all \(x \gt B\), then

\[\lim_{x \to +\infty} g(x) \le \lim_{x \to +\infty} f(x).\]

Proof

TODO

Infinite One-Sided Limits#

Definition: Negative Infinity as a Left-Sided Limit

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(c \in \mathbb{R}\) be an accumulation point of \(\mathcal{D} \cap (-\infty; c)\).

We say that \(-\infty\) is the left-sided limit of \(f\) at \(c\) if for each \(A \in \mathbb{R}\) there is a \(\delta \gt 0\) such that for all \(x \in \mathcal{D}\) with \(x \lt c\) we have:

\[0 \lt |x - c| \lt \delta \implies f(x) \lt A\]

Notation

\[\lim_{x\to c^-} f(x) = -\infty\]

Definition: Positive Infinity as a Left-Sided Limit

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(c \in \mathbb{R}\) be an accumulation point of \(\mathcal{D} \cap (-\infty; c)\).

We say that \(+\infty\) is the left-sided limit of \(f\) at \(c\) if for each \(A \in \mathbb{R}\) there is a \(\delta \gt 0\) such that for all \(x \in \mathcal{D}\) with \(x \lt c\) we have:

\[0 \lt |x-c| \lt \delta \implies f(x) \gt A\]

Notation

\[\lim_{x\to c^-} f(x) = \infty\]

Definition: Negative Infinity as a Right-Sided Limit

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(c \in \mathbb{R}\) be an accumulation point of \(\mathcal{D} \cap (c;+\infty)\).

We say that \(-\infty\) is the right-sided limit of \(f\) at \(c\) if for each \(A \in \mathbb{R}\) there is a \(\delta \gt 0\) such that for all \(x \in \mathcal{D}\) with \(x \gt c\) we have:

\[0 \lt |x - c| \lt \delta \implies f(x) \lt A\]

Notation

\[\lim_{x\to c^+} f(x) = -\infty\]

Definition: Positive Infinity as a Right-Sided Limit

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(c \in \mathbb{R}\) be an accumulation point of \(\mathcal{D} \cap (c;+\infty)\).

We say that \(+\infty\) is the right-sided limit of \(f\) at \(c\) if for each \(A \in \mathbb{R}\) there is a \(\delta \gt 0\) such that for all \(x \in \mathcal{D}\) with \(x \gt c\) we have:

\[0 \lt |x-c| \lt \delta \implies f(x) \gt A\]

Notation

\[\lim_{x\to c^+} f(x) = \infty\]

Infinite Limits#

Definition: Negative Infinity as Two-Sided Limit

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(c \in \mathbb{R}\) be an accumulation point of \(\mathcal{D}\).

We say that \(f\) approaches \(-\infty\) as \(x\) approaches \(c \in \mathbb{R}\) or that \(-\infty\) is the limit of \(f\) at \(c\) if for each \(A \in \mathbb{R}\) there is some \(\delta \gt 0\) such that

\[0 \lt |x-c| \lt \delta \implies f(x) \lt A\]

for all \(x \in \mathcal{D}\).

Notation

\[\lim_{x\to c} f(x) = -\infty\]

Definition: Negative Infinity as Limit at Negative Infinity

Let \(\mathcal{D}\) be unbounded below and let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

We say that \(f\) approaches \(-\infty\) as \(x\) approaches \(-\infty\) or that \(-\infty\) is the limit of \(f\) at \(-\infty\) if for each \(A \in \mathbb{R}\) there is some \(B \in \mathbb{R}\) such that \(f(x) \lt A\) for all \(x \in \mathcal{D}\) with \(x \lt B\).

Notation

\[\lim_{x \to -\infty} f(x) = -\infty\]

Definition: Negative Infinity as Limit at Positive Infinity

Let \(\mathcal{D}\) be unbounded above and let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

We say that \(f\) approaches \(-\infty\) as \(x\) approaches \(+\infty\) or that \(-\infty\) is the limit of \(f\) at \(+\infty\) if for each \(A \in \mathbb{R}\) there is some \(B \in \mathbb{R}\) such that \(f(x) \lt A\) for all \(x \in \mathcal{D}\) with \(x \gt B\).

Notation

\[\lim_{x \to +\infty} f(x) = -\infty\]

Definition: Positive Infinity as Two-Sided Limit

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(c \in \mathbb{R}\) be an accumulation point of \(\mathcal{D}\).

We say that \(f\) approaches \(+\infty\) as \(x\) approaches \(c \in \mathbb{R}\) or that \(+\infty\) is the limit of \(f\) at \(c\) if for each \(A \in \mathbb{R}\) there is some \(\delta \gt 0\) such that

\[0 \lt |x-c| \lt \delta \implies f(x) \gt A\]

for all \(x \in \mathcal{D}\).

Notation

\[\lim_{x\to c} f(x) = +\infty\]

Definition: Positive Infinity as Limit at Negative Infinity

Let \(\mathcal{D}\) be unbounded below and let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

We say that \(f\) approaches \(+\infty\) as \(x\) approaches \(-\infty\) or that \(+\infty\) is the limit of \(f\) at \(-\infty\) if for each \(A \in \mathbb{R}\) there is some \(B \in \mathbb{R}\) such that \(f(x) \gt A\) for all \(x \in \mathcal{D}\) with \(x \lt B\).

Notation

\[\lim_{x \to -\infty} f(x) = +\infty\]

Definition: Positive Infinity as Limit at Positive Infinity

Let \(\mathcal{D}\) be unbounded above and let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

We say that \(f\) approaches \(+\infty\) as \(x\) approaches \(+\infty\) or that \(+\infty\) is the limit of \(f\) at \(+\infty\) if for each \(A \in \mathbb{R}\) there is some \(B \in \mathbb{R}\) such that \(f(x) \gt A\) for all \(x \in \mathcal{D}\) with \(x \gt B\).

Notation

\[\lim_{x \to +\infty} f(x) = +\infty\]

Theorem: Arithmetic with Infinite Limits

Let \(f\) and \(g\) be real functions and let \(c \in \mathbb{R} \cup \{-\infty, +\infty\}\).

The following rules apply for the [limits](./Limits%20(Real%20Functions.md) of \(f\) and \(g\) at \(c\), either real or infinite:

	\(\displaystyle \lim_{x\to c} f(x) = L \lt 0\)	\(\displaystyle \lim_{x\to c} f(x) = L \gt 0\)
\(\displaystyle \lim_{x\to c} g(x) = -\infty\)	\(\displaystyle \lim_{x\to c} (f(x) + g(x)) = -\infty\) \(\displaystyle\lim_{x\to c} (f(x)g(x)) = +\infty\)	\(\displaystyle \lim_{x\to c} (f(x) + g(x)) = -\infty\) \(\displaystyle\lim_{x\to c} (f(x)g(x)) = -\infty\)
\(\displaystyle \lim_{x\to c} g(x) = + \infty\)	\(\displaystyle \lim_{x\to c} (f(x) + g(x)) = +\infty\) \(\displaystyle\lim_{x\to c} (f(x)g(x)) = -\infty\)	\(\displaystyle \lim_{x\to c} (f(x) + g(x)) = +\infty\) \(\displaystyle\lim_{x\to c} (f(x)g(x)) = +\infty\)

	\(\displaystyle \lim_{x\to c} f(x) = -\infty\)	\(\displaystyle \lim_{x\to c} f(x) = +\infty\)
\(\displaystyle \lim_{x\to c} g(x) = -\infty\)	\(\displaystyle \lim_{x\to c} (f(x) + g(x)) = -\infty\) \(\displaystyle\lim_{x\to c} (f(x)g(x)) = +\infty\)	\(\displaystyle \lim_{x\to c} (f(x) + g(x)) = \, ?\) \(\displaystyle\lim_{x\to c} (f(x)g(x)) = -\infty\)
\(\displaystyle \lim_{x\to c} g(x) = + \infty\)	\(\displaystyle \lim_{x\to c} (f(x) + g(x)) = \, ?\) \(\displaystyle\lim_{x\to c} (f(x)g(x)) = -\infty\)	\(\displaystyle \lim_{x\to c} (f(x) + g(x)) = +\infty\) \(\displaystyle\lim_{x\to c} (f(x)g(x)) = +\infty\)

Note

A question mark ("?") indicates that we cannot compute the limit directly, but we can try to transform the expression via algebraic manipulations in such a way, so as to make the limit computable.

Proof

TODO

Theorem: Converting between Limits at Infinity and Limits at Zero

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(L \in \mathbb{R} \cup \{-\infty, +\infty \}\).

The following rules allow us to convert between [limits at infinity](./Limits%20(Real%20Functions.md) and [one-sided limits](./Limits%20(Real%20Functions.md) at zero:

\[\lim_{x \to +\infty} f(x) = L \iff \lim_{y \to 0^+} f\left(\frac{1}{y}\right) = L\]

\[\lim_{x \to -\infty} f(x) = L \iff \lim_{y \to 0^-} f\left(\frac{1}{y}\right) = L\]

Proof

TODO

Theorem: Converting between Infinite Limits and Zero Limits

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(c \in \mathbb{R}\).

The following rules allow us to convert between infinite limits and real limits:

If \(\lim_{x \to c} f(x) = + \infty\) or \(\lim_{x \to c} f(x) = - \infty\), then \(\lim_{x \to c} \frac{1}{f(x)} = 0\), i.e.

\[\lim_{x \to c} f(x) = \pm \infty \implies \lim_{x \to c} \frac{1}{f(x)} = 0.\]

If \(\lim_{x \to c} f(x) = 0\) and there exists some neighborhood \(\mathcal{N}\) of \(c\) on which \(f(x) \gt 0\), then \(\lim_{x \to c} \frac{1}{f(x)} = + \infty\), i.e.

\[\lim_{x \to c} f(x) = 0 \text{ and } f(x) \gt 0 \text{ for all } x \in \mathcal{N} \implies \lim_{x \to c} \frac{1}{f(x)} = + \infty.\]

If \(\lim_{x \to c} f(x) = 0\) and neighborhood \(\mathcal{N}\) of \(c\) on which \(f(x) \lt 0\), then \(\lim_{x \to c} \frac{1}{f(x)} = - \infty\), i.e.

\[\lim_{x \to c} f(x) = 0 \text{ and } f(x) \lt 0 \text{ for all } x \in \mathcal{N} \implies \lim_{x \to c} \frac{1}{f(x)} = - \infty.\]

Proof

TODO

Theorem: Limits of Compositions

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be real functions and let \(c \in \mathbb{R}\).

If the limit of \(g\) exists at \(c\) and \(f\) is continuous at \(\lim_{x \to c} g(x)\), then the limit of their composition exists and

\[\lim_{x \to c} f(g(x)) = f\left(\lim_{x \to c} g(x) \right).\]

Proof

TODO

Theorem: L'Hôpital's Rule (Left-Sided Limits)

Let \(c \in \mathbb{R}\) and let \(f\) and \(g\) be real functions that are differentiable on an open interval \((a, c)\) (where \(a < c\)) with \(g'(x) \neq 0\) for all \(x \in (a, c)\).

If the limit of \(\lim_{x \to c^-} \frac{f'(x)}{g'(x)}\) is

\[\lim_{x \to c^-} \frac{f'(x)}{g'(x)} = L \in \mathbb{R} \cup \{-\infty, +\infty\}\]

and one of the following conditions holds:

zero limit condition: \(\lim_{x \to c^-} f(x) = 0\) and \(\lim_{x \to c^-} g(x) = 0\);
infinite limit condition: \(\lim_{x \to c^-} |g(x)| = +\infty\);

then the [limit](./Limits%20(Real%20Functions.md) \(\lim_{x \to c^-} \frac{f(x)}{g(x)}\) is equal to the [limit](./Limits%20(Real%20Functions.md) \(\lim_{x \to c^-} \frac{f'(x)}{g'(x)}\):

\[\lim_{x \to c^-} \frac{f(x)}{g(x)} = \lim_{x \to c^-} \frac{f'(x)}{g'(x)} = L\]

Proof

TODO

Theorem: L'Hôpital's Rule (Right-Sided Limits)

Let \(c \in \mathbb{R}\) and let \(f\) and \(g\) be real functions that are differentiable on an open interval \((c, b)\) (where \(b > c\)) with \(g'(x) \neq 0\) for all \(x \in (c, b)\).

If the limit of \(\lim_{x \to c^+} \frac{f'(x)}{g'(x)}\) is

\[\lim_{x \to c^+} \frac{f'(x)}{g'(x)} = L \in \mathbb{R} \cup \{-\infty, +\infty\}\]

and one of the following conditions holds:

zero limit condition: \(\lim_{x \to c^+} f(x) = 0\) and \(\lim_{x \to c^+} g(x) = 0\);
infinite limit condition: \(\lim_{x \to c^+} |g(x)| = +\infty\);

then the [limit](./Limits%20(Real%20Functions.md) \(\lim_{x \to c^+} \frac{f(x)}{g(x)}\) is equal to the [limit](./Limits%20(Real%20Functions.md) \(\lim_{x \to c^+} \frac{f'(x)}{g'(x)}\):

\[\lim_{x \to c^+} \frac{f(x)}{g(x)} = \lim_{x \to c^+} \frac{f'(x)}{g'(x)} = L\]

Proof

TODO

Example: \(\lim_{x \to 0^+} x \ln x\)

We want to determine the following [limit](./Limits%20(Real%20Functions.md):

\[\lim_{x \to 0^+} x \ln x\]

We first transform the expression:

\[\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x}} = \lim_{x \to 0^+} \frac{f(x)}{g(x)}\]

We see that \(\lim_{x \to 0^+} |g(x)| = \infty\), so this is a candidate for [L'Hôpital's rule](./Limits%20(Real%20Functions.md). By differentiating, we obtain the following:

\[f'(x) = \frac{1}{x} \qquad g'(x) = -\frac{1}{x^2}\]

We see that \(g'(x) \ne 0\) for all \(x \in \mathbb{R} \setminus \{0\}\). Furthermore:

\[\lim_{x \to 0^+} \frac{f'(x)}{g'(x)} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} (-x) = 0 \in \mathbb{R} \cup \{-\infty, +\infty\}\]

Therefore, we can use [L'Hôpital's rule](./Limits%20(Real%20Functions.md):

\[\lim_{x \to 0^+} \frac{f(x)}{g(x)} = \lim_{x \to 0^+} \frac{f'(x)}{g'(x)} = 0\]

Theorem: L'Hôpital's Rule (Two-Sided Limits and Limits at \(\pm \infty\))

Let \(c \in \mathbb{R} \cup \{-\infty, +\infty\}\) and let \(f\) and \(g\) be real functions that are differentiable on a deleted neighborhood \(\mathcal{N}(c)\) with \(g'(x) \neq 0\) for all \(x \in \mathcal{N}(c)\).

If the limit of \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\) is

\[\lim_{x \to c} \frac{f'(x)}{g'(x)} = L \in \mathbb{R} \cup \{-\infty, +\infty\}\]

and one of the following conditions holds:

zero limit condition: \(\lim_{x \to c} f(x) = 0\) and \(\lim_{x \to c} g(x) = 0\);
infinite limit condition: \(\lim_{x \to c} |g(x)| = +\infty\);

then the [limit](./Limits%20(Real%20Functions.md) \(\lim_{x \to c} \frac{f(x)}{g(x)}\) is equal to the [limit](./Limits%20(Real%20Functions.md) \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\):

\[\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} = L\]

Proof

TODO

Example: \(\lim_{x \to 0} \frac{\mathrm{e}^x - 1}{x}\)

We want to determine the following [limit](./Limits%20(Real%20Functions.md):

\[\lim_{x \to 0} \frac{\mathrm{e}^x - 1}{x}\]

We see that \(\lim_{x \to 0} (\mathrm{e}^{x} - 1) = \lim_{x \to 0} x = 0\), so we might be able to use [L'Hôpital's rule](./Limits%20(Real%20Functions.md) provided that the appropriate conditions are met. We see first that \((x)' = 1\) and so \((x') \ne 0\) for all \(x \in \mathbb{R}\). We see that \(\lim_{x \to 0} \frac{(\mathrm{e}^x - 1)'}{(x)'} \in \mathbb{R} \cup \{-\infty, +\infty\}\):

\[\lim_{x \to 0} \frac{(\mathrm{e}^x - 1)'}{(x)'} = \lim_{x \to 0} \frac{\mathrm{e}^x}{1} = 1\]

Therefore:

\[\lim_{x \to 0} \frac{\mathrm{e}^x - 1}{x} = \lim_{x \to 0} \frac{(\mathrm{e}^x - 1)'}{(x)'} = 1\]

Example: \(\lim_{x \to 0} \left(\frac{1}{\sin x} - \frac{1}{x}\right)\)

We want to determine the following [limit](./Limits%20(Real%20Functions.md):

\[\lim_{x \to 0} \left(\frac{1}{\sin x} - \frac{1}{x}\right)\]

We first transform the expression a bit:

\[\frac{1}{\sin x} - \frac{1}{x} = \frac{x - \sin x}{x \sin x}\]

Let \(f(x) = x - \sin x\) and \(g(x) = x \sin x\). We immediately see that \(\lim_{x \to 0} f(x) = \lim_{x \to 0} g(x) = 0\), so we might be able to use [L'Hôpital's rule](./Limits%20(Real%20Functions.md). By differentiating, we obtain the following:

\[f'(x) = 1 - \cos x \qquad g'(x) = \sin x + x \cos x\]

We can see that \(g'(x) \ne 0\) for all \(x \in (-\frac{\pi}{2}; 0)\cup (0; +\frac{\pi}{2})\).

However, we also see that \(\lim_{x \to 0} f'(x) = \lim_{x \to 0} g'(x) = 0\), so we cannot use [L'Hôpital's rule](./Limits%20(Real%20Functions.md) directly. Nevertheless, we might be able to use apply use it for \(f'\) and \(g'\) as well. By differentiating again, we obtain the following:

\[f''(x) = \sin x \qquad g''(x) = 2 \cos x - x \sin x\]

We see that \(g''(x) \ne 0\) for all \(x \in (-\frac{\pi}{4};0)\cup (0; +\frac{\pi}{4})\). Moreover, we have:

\[\lim_{x \to 0} \frac{f''(x)}{g''(x)} = \lim_{x \to 0}\frac{\sin x}{2 \cos x - x \sin x} = \frac{0}{2} = 0\]

We can therefore use [L'Hôpital's rule](./Limits%20(Real%20Functions.md):

\[\lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{f''(x)}{g''(x)} = 0\]

\[\lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} = 0\]

Theorem: Zero Limit via Boundedness

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be real functions and let \(c \in \mathbb{R} \cup \{-\infty, +\infty\}\).

If there exists a deleted neighborhood of \(c\) (an interval of the form \((-\infty, M)\) or \((M, +\infty)\) for \(c = \pm \infty\)) on which \(f\) is bounded and the limit of \(g\) at \(c\) is zero, then

\[\lim_{x \to c} (g(x) f(x)) = 0.\]

Proof

TODO

Theorem: Important Trigonometric Limits

Following are some [limits](./Limits%20(Real%20Functions.md) involving the real trigonometric functions:

\[\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{x}{\sin x} = 1\]

\[\lim_{x \to 0} \frac{1 - \cos x}{x} = 0\]

Proof

TODO

Theorem: Important Exponential Limits

Following are some [limits](./Limits%20(Real%20Functions.md) involving the real exponential function:

\[\lim_{x \to -\infty} \left(1 + \frac{1}{x}\right)^x = \lim_{x \to +\infty} \left(1 + \frac{1}{x}\right)^x = \mathrm{e}\]

\[\lim_{x \to 0} (1 + x)^{\frac{1}{x}} = \mathrm{e}\]

Proof

TODO

Asymptotes#

Definition: Vertical Asymptote

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

We say that \(f\) has a vertical asymptote at \(x = c \in \mathbb{R}\) if it has at least one infinite one-sided limit, i.e. at least one of the following holds:

\(\displaystyle \lim_{x \to c^-} f(x) = -\infty\)
\(\displaystyle \lim_{x \to c^+} f(x) = -\infty\)
\(\displaystyle \lim_{x \to c^-} f(x) = \infty\)
\(\displaystyle \lim_{x \to c^+} f(x) = \infty\)

Intuition

Intuitively, this definition means that the value of \(f(x)\) gets closer and closer to \(- \infty\) or \(+ \infty\) as \(x\) approaches \(c\) either from the left or from the right.

Definition: Horizontal Asymptote

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

We say that \(f\) has a horizontal asymptote \(y = a \in \mathbb{R}\) if the limit of \(f\) as \(x\) approaches positive or negative infinity is \(a\), i.e. if at least one of the following holds:

\(\displaystyle \lim_{x \to -\infty} f(x) = a\)
\(\displaystyle \lim_{x \to \infty} f(x) = a\)

Intuition

Intuitively, this definition means that the value of \(f(x)\) gets closer and closer to \(a\) as \(x\) approaches either positive or negative infinity.

Definition: Oblique Asymptote

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

The line \(y = ax + b\) is an oblique or slanted asymptote of \(f\) if at least one of the following is true:

\(\displaystyle \lim_{x \to -\infty} [f(x) - (ax + b)] = 0\)
\(\displaystyle \lim_{x \to \infty} [f(x) - (ax + b)] = 0\)

Intuition

Intuitively, this definition means \(f(x)\) gets closer and closer to the line \(y = ax + b\) as \(x\) approaches either positive or negative infinity.

Theorem: Oblique Asymptotes

Let \(f\) be a real function.

The line \(y = ax + b\) is an asymptote of \(f\) if and only if the limits \(\lim_{x \to \pm \infty} \frac{f(x)}{x}\) and \(\lim_{x \to \pm \infty} (f(x) - ax)\) exist and

\[\begin{aligned} a &= \lim_{x \to \pm \infty} \frac{f(x)}{x} \\ \\ b &= \lim_{x \to \pm \infty} (f(x) - ax)\end{aligned}\]

Proof

TODO