3.1: Limits of Functions

Definition \(\PageIndex{1}\)

Let \(f: D \rightarrow \mathbb{R}\) and let \(\bar{x}\) be a limit point of \(D\). We say that \(f\) has a limit at \(\bar{x}\) if there exists a real number \(\ell\) such that for every \(\varepsilon>0\), there exists \(\delta>0\) with

\[|f(x)-\ell|<\varepsilon\]

for all \(x \in D\) for which \(0<|x-\bar{x}|<\delta\). In this case, we write

\[\lim _{x \rightarrow \bar{x}} f(x)=\ell.\]

Remark \(\PageIndex{1}\)

Note that the limit point \(\bar{x}\) in the definition of limt may or may not be an element of the domain \(D\). In any case, the inequality \(|f(x)-\ell|<\varepsilon\) need only be satisfied by elements of \(D\).

Example \(\PageIndex{1}\)

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by \(f(x)=5 x-7\). We prove that \(\lim _{x \rightarrow 2} f(x)=3\).

Solution

Let \(\varepsilon>0\). First note that \(|f(x)-2|=|5 x-7-3|=|5 x-10|=5|x-2|\). This suggests the choice \(\delta=\varepsilon / 5\). Then, if \(|x-2|<\delta\) we have

\[|f(x)-2|=5|x-2|<5 \delta=\varepsilon. \nonumber\]

Example \(\PageIndex{2}\)

Let \(f:[0,1) \rightarrow \mathbb{R}\) be given by \(f(x)=x^{2}+x\). Let \(\bar{x} =1\) and \(\ell = 2\).

Solution

First note that \(|f(x)-\ell|=\left|x^{2}+x-2\right|=|x-1||x+2|\) and for \(x \in[0,1)\). \(|x+2| \leq|x|+2 \leq 3\). Now, given \(\varepsilon>0\), choose \(\delta=\varepsilon / 3\). Then, if \(|x-1|<\delta\) and \(x \in[0,1)\), we have

\[|f(x)-\ell|=\left|x^{2}+x-2\right|=|x-1||x+2|<3 \delta=3 \frac{\varepsilon}{3}=\varepsilon. \nonumber\]

This shows that \(\lim _{x \rightarrow 1} f(x)=2\).

Example \(\PageIndex{3}\)

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by \(f(x)=x^{2}\). We show that \(\lim _{x \rightarrow 2} f(x)=4\).

Solution

First note that \(|f(x)-4|=\left|x^{2}-4\right|=|(x-2)(x+2)|=|x-2||x+2|\). Since the domain is all of \(\mathbb{R}\) the expression \(|x+2|\) is not bounded and we cannot proceed as in Example \(\PageIndex{2}\). However, we are interested only in values of \(x\) close to \(2\) and, thus, we impose the condition \(\delta \leq 1\). If \(|x-2|<1\), then \(-1<x-2<1\) and, so, \(1<x<3\). It follows, for such \(x\), that \(|x|<3\) and, hence \(|x|+2<5\).

Now, given \(\varepsilon>0\) we choose \(\delta=\min \left\{1, \frac{\varepsilon}{5}\right\}\). Then, whenever \(|x-2|<\delta\) we get

\[|f(x)-4|=|x-2| x+2|\leq| x-2|(|x|+2)<\delta 5 \leq \varepsilon. \nonumber\]

Example \(\PageIndex{4}\)

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by \(f(x)=\frac{3 x-5}{x^{2}+3}\). We prove that \(\lim _{x \rightarrow 1} f(x)=-\frac{1}{2}\).

Solution

First we look at the expression \(\left|f(x)-\left(-\frac{1}{2}\right)\right|\) and try to identify a factor \(|x-1|\) (because here \(\bar{x} = 1\).

\[\left|f(x)-\left(-\frac{1}{2}\right)\right|=\left|\frac{3 x-5}{x^{2}+3}+\frac{1}{2}\right|=\left|\frac{6 x-10+x^{2}+3}{x^{2}+3}\right|=\frac{|x-1||x+7|}{2\left(x^{2}+3\right)} \leq \frac{1}{6}|x-1||x+7|. \nonumber\]

Proceeding as in the previous example, if \(|x-1|<1\) we get \(-1<x-1<1\) and, so, \(0<x<2\). Thus \(|x|<2\) and \(|x+7| \leq|x|+2<9\).

Now, given \(\varepsilon>0\), we choose \(\delta=\min \left\{1, \frac{2}{3} \varepsilon\right\}\). It follows that if \(|x-1|<\delta\) we get

\[\left|f(x)-\left(-\frac{1}{2}\right)\right| \leq \frac{|x+7|}{6}|x-1|<\frac{9}{6} \delta \leq \varepsilon. \nonumber\]

The following theorem will let us apply our earlier results on limits of sequences to obtain new results on limits of functions.

Theorem \(\PageIndex{2}\): Sequential Characterization of Limits

Let \(f: D \rightarrow \mathbb{R}\) and let \(\bar{x}\) be a limit point of \(D\). Then

\[\lim _{x \rightarrow \bar{x}} f(x)=\ell\]

if and only if

\[\lim _{n \rightarrow \infty} f\left(x_{n}\right)=\ell\]

for every sequence \(\left\{x_{n}\right\}\) in \(D\) such that \(x_{n} \neq \bar{x}\) for every \(n\) and \(\left\{x_{n}\right\}\) converges to \(\bar{x}\).

Proof

Suppose (3.1) holds. Let \(\left\{x_{n}\right\}\) be a sequence in \(D\) with \(x_{n} \neq \bar{x}\) for every \(n\) and such that \(\left\{x_{n}\right\}\) converges to \(\bar{x}\). given any \(\varepsilon>0\), there exists \(\delta > 0\) such that \(|f(x)-\ell|<\varepsilon\) whenever \(x \in D\) and \(0<|x-\bar{x}|<\delta\). Then there exists \(N \in \mathbb{N}\) with \(0<\left|x_{n}-\bar{x}\right|<\delta\) for all \(n \geq N\). For such \(n\), we have

\[\left|f\left(x_{n}\right)-\ell\right|<\varepsilon.\]

This implies (3.2).

Conversely, suppose (3.1) is false. Then there exists \(\varepsilon_{0}>0\) such that for every \(\delta > 0\), there exists \(x \in D\) with \(0<|x-\bar{x}|<\delta\) and \(|f(x)-\ell| \geq \varepsilon_{0}\). Thus, for every \(n \in \mathbb{N}\), there exits \(x_{n} \in D\) with \(0<\left|x_{n}-\bar{x}\right|<\frac{1}{n}\) and \(\left|f\left(x_{n}\right)-\ell\right| \geq \varepsilon_{0}\). By the squeeze theorem (Theorem 2.1.6), the sequence \(\left\{x_{n}\right\}\) converges to \(\bar{x}\). Moreover, \(x_{n} \neq \bar{x}\) for every \(n\). This shows that (3.2) is false. It follows that (3.2) implies (3.1) and the proof is complete.

\(\square\)

Corollary \(\PageIndex{3}\)

Let \(f: D \rightarrow \mathbb{R}\) and let \(\bar{x}\) be a limit point of \(D\). \(f\) has a limit at \(\bar{x}\), then this limit is unique.

Proof

Suppose by contradiction that \(f\) has two different limits \(\ell_{1}\) and \(\ell_{2}\). \(\left\{x_{n}\right\}\) be a sequence in \(D \backslash\{\bar{x}\}\) that converges to \(\bar{x}\). By Theorem 3.1.2, the sequence \(\left\{f\left(x_{n}\right)\right\}\) converges to two different limits \(\ell_{1}\) and \(\ell_{2}\). this is a contradiction to Theorem 2.1.3. \(\square\)

The following corollary follows directly from Theorem 3.1.2.

Corollary \(\PageIndex{4}\)

Let \(f: D \rightarrow \mathbb{R}\) and let \(\bar{x}\) be a limit point of \(D\). Then \(f\) does not have a limit at \(\bar{x}\) if and only if there exists a sequence \(\left\{x_{n}\right\}\) in \(D\) such that \(x_{n} \neq \bar{x}\) for every \(n\), \(\left\{x_{n}\right\}\) converges to \(\bar{x}\), and \(\left\{f\left(x_{n}\right)\right\}\) does not converge.

Proof

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Example \(\PageIndex{5}\)

Consider the Dirichlet function \(f: \mathbb{R} \rightarrow \mathbb{R}\) given by

\[f(x)=\left\{\begin{array}{ll}
1, & \text { if } x \in \mathbb{Q} \\
0, & \text { if } x \in \mathbb{Q}^{c}
\end{array}\right .\]

Then \(\lim _{x \rightarrow \bar{x}} f(x)\) does not exist for any \(\bar{x} \in \mathbb{R}\).

Solution

Indeed, fix \(\bar{x} \in \mathbb{R}\) and choose two sequences \(\left\{r_{n}\right\}\), \(\left\{s_{n}\right\}\) converging to \(\bar{x}\) such that \(r_{n} \in \mathbb{Q}\) and \(s_{n} \notin \mathbb{Q}\) for all \(n \in \mathbb{N}\). Define a new sequence \(\left\{x_{n}\right\}\) by

\[x_{n}=\left\{\begin{array}{ll}
r_{k}, & \text { if } n=2 k \\
s_{k}, & \text { if } n=2 k-1
\end{array}\right .\]

It is clear that \(\left\{x_{n}\right\}\) converges to \(\bar{x}\). Moreover, since \(\left\{f\left(r_{n}\right)\right\}\) converges to \(1\) and \(\left\{f\left(s_{n}\right)\right\}\) converges to \(0\). Theorem 2.1.9 implies that the sequence \(\left\{f\left(x_{n}\right)\right\}\) does not converge. It follows from the sequential characterization of limits that \(\lim _{x \rightarrow \bar{x}} f(x)\) does not exists.

Theorem \(\PageIndex{5}\)

Let \(f, g: D \rightarrow \mathbb{R}\) and let \(\bar{x}\) be a limit point of \(D\). Suppose that

\[\lim _{x \rightarrow \bar{x}} f(x)=\ell_{1}, \lim _{x \rightarrow \bar{x}} g(x)=\ell_{2} ,\]

and that there exists \(\delta>0\) such that

\[f(x) \leq g(x) \text { for all } x \in B(\bar{x} ; \delta) \cap D, x \neq \bar{x} .\]

Then \(\ell_{1} \leq \ell_{2}\)

Proof

Let \(\left\{x_{n}\right\}\) be a sequence in \(B(\bar{x} ; \delta) \cap D=(\bar{x}-\delta, \bar{x}+\delta) \cap D\) that converges to \(\bar{x}\) and \(x_{n} \neq \bar{x}\) for all \(n\). By Theorem 3.1.2,

\[\lim _{n \rightarrow \infty} f\left(x_{n}\right)=\ell_{1} \text { and } \lim _{n \rightarrow \infty} g\left(x_{n}\right)=\ell_{2} .\]

Since \(f\left(x_{n}\right) \leq g\left(x_{n}\right)\) for all \(n \in \mathbb{N}\), applyig Theorem 2.1.5, we obtain \(\ell_{1} \leq \ell_{2}\). \(\square\)

Theorem \(\PageIndex{6}\)

Let \(f, g: D \rightarrow \mathbb{R}\) and let \(\bar{x}\) be a limit point of \(D\). Suppose

\[\lim _{x \rightarrow \bar{x}} f(x)=\ell_{1}, \lim _{x \rightarrow \bar{x}} g(x)=\ell_{2} ,\]

and \(\ell_{1}<\ell_{2}\). Then there exists \(\delta > 0\) such that

\[f(x)<g(x) \text { for all } x \in B(\bar{x} ; \delta) \cap D, x \neq \bar{x} .\]

Proof

Choose \(\varepsilon>0\) such that \(\ell_{1}+\varepsilon<\ell_{2}-\varepsilon\) (equivalently, such that \(\varepsilon<\frac{\ell_{2}-\ell_{1}}{2}\)). Then there exists \(\delta > 0\) such that

\[\ell_{1}-\varepsilon<f(x)<\ell_{1}+\varepsilon \text { and } \ell_{2}-\varepsilon<g(x)<\ell_{2}+\varepsilon\]

for all \(x \in B(\bar{x} ; \delta) \cap D, x \neq \bar{x}\). Thus,

\[f(x)<\ell_{1}+\varepsilon<\ell_{2}-\varepsilon<g(x) \text { for all } x \in B(\bar{x} ; \delta) \cap D, x \neq \bar{x} .\]

The proof is now complete. \(\square\)

Theorem \(\PageIndex{7}\)

Let \(f, g, h: D \rightarrow \mathbb{R}\) and let \(\bar{x}\) be a limit point of \(D\). Suppose there exists \(\delta > 0\) such that \(f(x) \leq g(x) \leq h(x)\) for all \(x \in B(\bar{x} ; \delta) \cap D, x \neq \bar{x}\). If \(\lim _{x \rightarrow \bar{x}} f(x)=\lim _{x \rightarrow \bar{x}} h(x)=\ell\), then \(\lim _{x \rightarrow \bar{x}} g(x)=\ell\).

Proof

The proof is straightforward using Theorem 2.1.6 and Theorem 3.1.2. \(\square\)

Remark \(\PageIndex{8}\)

We will adopt the following convention. When we write \(\lim _{x \rightarrow \bar{x}} f(x)\) without specifying the domain \(D\) of \(f\) we will assume the \(D\) is the largest subset of \(\mathbb{R}\) such that if \(x \in D\), then \(f(x)\) results in a real number. For example, in

\[\lim _{x \rightarrow 2} \frac{1}{x+3}\]

we assume \(D=\mathbb{R} \backslash\{-3\}\) and in

\[\lim _{x \rightarrow 1} \sqrt{x}\]

we assume \(D=[0, \infty)\).

Exercise \(\PageIndex{1}\)

Use the definition of limit to prove that

1. \(\lim _{x \rightarrow 2} 3 x-7=-1\)
2. \(\lim _{x \rightarrow 3}\left(x^{2}+1\right)=10\).
3. \(\lim _{x \rightarrow 1} \frac{x+3}{x+1}=2\).
4. \(\lim _{x \rightarrow 0} \sqrt{x}=0\).
5. \(\lim _{x \rightarrow 2} x^{3}=8\).

Answer

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Exercise \(\PageIndex{2}\)

Prove that the following limits do not exist.

1. \(\lim _{x \rightarrow 0} \frac{x}{|x|}\).
2. \(\lim _{x \rightarrow 0} \cos (1 / x)\).

Answer

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Exercise \(\PageIndex{3}\)

Let \(f: D \rightarrow \mathbb{R}\) and let \(\bar{x}\) be a limit point of \(D\). Prove that if \(\lim _{x \rightarrow \bar{x}} f(x)=\ell\), then

\[\lim _{x \rightarrow \bar{x}}|f(x)|=|\ell| .\]

Give an example to show that the convers is not true in general.

Answer

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Exercise \(\PageIndex{4}\)

Let \(f: D \rightarrow \mathbb{R}\) and let \(\bar{x}\) be a limit point of \(D\). Suppose \(f(x) \geq 0\) for all \(x \in D\). Prove that if \(\lim _{x \rightarrow \bar{x}} f(x)=\ell\), then

\[\lim _{x \rightarrow \bar{x}} \sqrt{f(x)}=\sqrt{\ell} .\]

Answer

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Exercise \(\PageIndex{5}\)

Find \(\lim _{x \rightarrow 0} x \sin (1 / x)\).

Answer

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Exercise \(\PageIndex{6}\)

Let \(f\) be the function given by

\[f(x)=\left\{\begin{array}{ll}
x, & \text { if } x \in \mathbb{Q} \cap[0,1] \text{;} \\
1-x, & \text { if } x \in \mathbb{Q}^{c} \cap[0,1] \text{.}
\end{array}\right .\]

Determine which of the following limits exist. For those that exist find their values.

1. \(\lim _{x \rightarrow 1 / 2} f(x)\).
2. \(\lim _{x \rightarrow 0} f(x)\).
3. \(\lim _{x \rightarrow 1} f(x)\).

Answer

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