The Limit of a Function

Whenever a point $x$ is within $\delta$ units of $c$ , $f(x)$ is within $\varepsilon$ units of $L$

In preliminary calculus, the concept of a limit is probably the most difficult one to grasp (after all, it took mathematicians 150 years to arrive at it); it is also the most important and most useful one.

The intuitive definition of a limit is inadequate to prove anything rigorously about it. The problem lies in the vague term "arbitrarily close". We discussed earlier that the meaning of this term is that the closer $x$ gets to the specified value, the closer the function must get to the limit, so that however close we want the function to the limit, we can accomplish this by making $x$ sufficiently close to our value. We can express this requirement technically as follows:

Formal definition of a limit

Let $f(x)$ be a function defined on an open interval $D$ that contains $c$ , except possibly at $x=c$ . Let $L$ be a number. Then we say that

$\lim _{x\to c}f(x)=L$

if, for every $\varepsilon >0$ , there exists a $\delta >0$ such that for all $x\in D$ with

$0<|x-c|<\delta$

we have

${\Big |}f(x)-L{\Big |}<\varepsilon$

To further explain, earlier we said that "however close we want the function to the limit, we can find a corresponding $x$ close to our value." Using our new notation of epsilon ($\varepsilon$) and delta ($\delta$), we mean that if we want to make $f(x)$ within $\varepsilon$ of $L$ , the limit, then we know that making $x$ within $\delta$ of $c$ puts it there.

Again, since this is tricky, let's resume our example from before: $f(x)=x^{2}$ , at $x=2$. To start, let's say we want $f(x)$ to be within .01 of the limit. We know by now that the limit should be 4, so we say: for $\varepsilon =0.01$ , there is some $\delta$ so that as long as $0<|x-c|<\delta$ , then ${\Big |}f(x)-L{\Big |}<\varepsilon$ .

To show this, we can pick any $\delta$ that is bigger than 0, so long as it works. For example, you might pick $10^{-14}$ , because you are absolutely sure that if $x$ is within $10^{-14}$ of 2, then $f(x)$ will be within $0.01$ of 4. This $\delta$ works for $\varepsilon =0.01$ . But we can't just pick a specific value for $\varepsilon$ , like 0.01, because we said in our definition "for every $\varepsilon >0$ ." This means that we need to be able to show an infinite number of $\delta$s, one for each $\varepsilon$ . We can't list an infinite number of $\delta$s!

Of course, we know of a very good way to do this; we simply create a function, so that for every $\varepsilon$ , it can give us a $\delta$ . In this case, one definition of $\delta$ that works is $\delta (\varepsilon )=\left\{{\begin{matrix}2{\sqrt {2}}-2&{\mbox{if }}\epsilon \geq 4\\{\sqrt {\epsilon +4}}-2&{\mbox{if }}\epsilon <4\end{matrix}}\right.$

So, in general, how do you show that $f(x)$ tends to $L$ as $x$ tends to $c$? Well imagine somebody gave you a small number $\varepsilon$ (e.g., say $\varepsilon =0.03$). Then you have to find a $\delta >0$ and show that whenever $0<|x-c|<\delta$ we have ${\Big |}f(x)-L{\Big |}<0.03$ . Now if that person gave you a smaller $\varepsilon$ (say $\varepsilon =0.002$) then you would have to find another $\delta$, but this time with 0.03 replaced by 0.002. If you can do this for any choice of $\varepsilon$ then you have shown that $f(x)$ tends to $L$ as $x$ tends to $c$ . Of course, the way you would do this in general would be to create a function giving you a $\delta$ for every $\varepsilon$ , just as in the example above.

Formal Definition of the Limit at Infinity

We call $L$ the limit of $f(x)$ as $x$ approaches $\infty$ if for every number $\varepsilon >0$ there exists a $\delta$ such that whenever $x>\delta$ we have

${\Big |}f(x)-L{\Big |}<\varepsilon$

When this holds we write

$\lim _{x\to \infty }f(x)=L$

or

$f(x)\to L$ as $x\to \infty$

Similarly, we call $L$ the limit of $f(x)$ as $x$ approaches $-\infty$ if for every number $\varepsilon >0$ , there exists a number $\delta$ such that whenever $x<\delta$ we have

${\Big |}f(x)-L{\Big |}<\varepsilon$

When this holds we write

$\lim _{x\to -\infty }f(x)=L$

or

$f(x)\to L$ as $x\to -\infty$

Notice the difference in these two definitions. For the limit of $f(x)$ as $x$ approaches $\infty$ we are interested in those $x$ such that $x>\delta$ . For the limit of $f(x)$ as $x$ approaches $-\infty$ we are interested in those $x$ such that $x<\delta$ .

Examples

Here are some examples of the formal definition.

Example 1

We know from earlier in the chapter that

$\lim _{x\to 8}{\frac {x}{4}}=2$

What is $\delta$ when $\varepsilon =0.01$ for this limit?

We start with the desired conclusion and substitute the given values for $f(x)$ and $\varepsilon$ :

$\left|{\frac {x}{4}}-2\right|<0.01$

Then we solve the inequality for $x$ :

$7.96<x<8.04$

This is the same as saying

$-0.04<x-8<0.04$

(We want the thing in the middle of the inequality to be $x-8$ because that's where we're taking the limit.) We normally choose the smaller of $|-0.04|$ and $0.04$ for $\delta$, so $\delta =0.04$ , but any smaller number will also work.

Example 2

What is the limit of $f(x)=x+7$ as $x$ approaches 4?

There are two steps to answering such a question; first we must determine the answer — this is where intuition and guessing is useful, as well as the informal definition of a limit — and then we must prove that the answer is right.

In this case, 11 is the limit because we know $f(x)=x+7$ is a continuous function whose domain is all real numbers. Thus, we can find the limit by just substituting 4 in for $x$ , so the answer is $4+7=11$ .

We're not done, though, because we never proved any of the limit laws rigorously; we just stated them. In fact, we couldn't have proved them, because we didn't have the formal definition of the limit yet, Therefore, in order to be sure that 11 is the right answer, we need to prove that no matter what value of $\varepsilon$ is given to us, we can find a value of $\delta$ such that

${\Big |}f(x)-11{\Big |}<\varepsilon$

whenever

$|x-4|<\delta$

For this particular problem, letting $\delta =\varepsilon$ works. Now, we have to prove

${\Big |}f(x)-11{\Big |}<\varepsilon$

given that

$|x-4|<\delta =\varepsilon$

Since $|x-4|<\varepsilon$ , we know

${\Big |}f(x)-11{\Big |}={\Big |}x+7-11{\Big |}=|x-4|<\varepsilon$

which is what we wished to prove.

Example 3

What is the limit of $f(x)=x^{2}$ as $x$ approaches 4?

As before, we use what we learned earlier in this chapter to guess that the limit is $4^{2}=16$ . Also as before, we pull out of thin air that

$\delta ={\sqrt {\varepsilon +16}}-4$

Note that, since $\varepsilon$ is always positive, so is $\delta$ , as required. Now, we have to prove

${\Big |}x^{2}-16{\Big |}<\varepsilon$

given that

$|x-4|<\delta ={\sqrt {\varepsilon +16}}-4$ .

We know that

$|x+4|={\Big |}(x-4)+8{\Big |}\leq |x-4|+8<\delta +8$

(because of the triangle inequality), so

${\begin{matrix}{\Big |}x^{2}-16{\Big |}&=&|x-4|\cdot |x+4|\\\\\ &<&\delta (\delta +8)\\\\\ &<&({\sqrt {16+\varepsilon }}-4)({\sqrt {16+\varepsilon }}+4)\\\\\ &<&({\sqrt {16+\varepsilon }})^{2}-4^{2}\\\\\ &=&\varepsilon +16-16\\\\\ &<&\varepsilon \end{matrix}}$

Example 4

Show that the limit of $\sin \left({\tfrac {1}{x}}\right)$ as $x$ approaches 0 does not exist.

We will proceed by contradiction. Suppose the limit exists; call it $L$ . For simplicity, we'll assume that $L\neq 1$ ; the case for $L=1$ is similar. Choose $\varepsilon =|1-L|$ . Then if the limit were $L$ there would be some $\delta >0$ such that $\left|\sin \left({\tfrac {1}{x}}\right)-L\right|<\varepsilon =|1-L|$ for every $x$ with $0<|x|<\delta$ . But, for every $\delta >0$ , there exists some (possibly very large) $n$ such that $0<x_{0}={\frac {1}{{\frac {\pi }{2}}+2\pi n}}<\delta$ , but $\left|\sin \left({\tfrac {1}{x_{0}}}\right)-L\right|=|1-L|$ , a contradiction.

Example 5

What is the limit of $x\sin \left({\tfrac {1}{x}}\right)$ as $x$ approaches 0?

By the Squeeze Theorem, we know the answer should be 0. To prove this, we let $\delta =\varepsilon$ . Then for all $x$ , if $0<|x|<\delta$ , then $\left|x\sin \left({\tfrac {1}{x}}\right)-0\right|\leq |x|<\varepsilon$ as required.

Example 6

Suppose that $\lim _{x\to a}f(x)=L$ and $\lim _{x\to a}g(x)=M$ . What is $\lim _{x\to a}{\big [}f(x)+g(x){\big ]}$ ?

Of course, we know the answer should be $L+M$ , but now we can prove this rigorously. Given some $\varepsilon$ , we know there's a $\delta _{1}$ such that, for any $x$ with $0<|x-a|<\delta _{1}$ , ${\Big |}f(x)-L{\Big |}<{\frac {\varepsilon }{2}}$ (since the definition of limit says "for any $\varepsilon$", so it must be true for ${\frac {\varepsilon }{2}}$ as well). Similarly, there's a $\delta _{2}$ such that, for any $x$ with $0<|x-a|<\delta _{2}$ , ${\Big |}g(x)-M{\Big |}<{\frac {\varepsilon }{2}}$ . We can set $\delta$ to be the lesser of $\delta _{1}$ and $\delta _{2}$ . Then, for any $x$ with $0<|x-a|<\delta$ , ${\bigg |}{\big (}f(x)+g(x){\big )}-(L+M){\bigg |}\leq {\Big |}f(x)-L{\Big |}+{\Big |}g(x)-M{\Big |}<{\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2}}=\varepsilon$ , as required.

If you like, you can prove the other limit rules too using the new definition. Mathematicians have already done this, which is how we know the rules work. Therefore, when computing a limit from now on, we can go back to just using the rules and still be confident that our limit is correct according to the rigorous definition.

Formal Definition of a Limit Being Infinity

Let $f(x)$ be a function defined on an open interval $D$ that contains $c$ , except possibly at $x=c$ . Then we say that

$\lim _{x\to c}f(x)=\infty$

if, for every $\varepsilon$ , there exists a $\delta >0$ such that for all $x\in D$ with

$0<|x-c|<\delta$

we have

$f(x)>\varepsilon$ .

When this holds we write

$\lim _{x\to c}f(x)=\infty$

or

$f(x)\to \infty$ as $x\to c$

Similarly, we say that

$\lim _{x\to c}f(x)=-\infty$

if, for every $\varepsilon$ , there exists a $\delta >0$ such that for all $x\in D$ with

$0<|x-c|<\delta$

we have

$f(x)<\varepsilon$ .

When this holds we write

$\lim _{x\to c}f(x)=-\infty$

or

$f(x)\to -\infty$ as $x\to c$.

Exact Integrals as Limits of Sums

Using the definition of an integral, we can evaluate the limit as $n$ goes to infinity. This technique requires a fairly high degree of familiarity with summation identities. This technique is often referred to as evaluation "by definition," and can be used to find definite integrals, as long as the integrands are fairly simple. We start with definition of the integral:

$\int \limits _{a}^{b}f(x)dx$	$=\lim _{n\to \infty }\left[{\frac {b-a}{n}}\sum _{k=1}^{n}f(x_{k}^{*})\right]$	Then picking $x_{k}^{*}$ to be $x_{k}=a+k{\frac {b-a}{n}}$ we get,
	$=\lim _{n\to \infty }\left[{\frac {b-a}{n}}\sum _{k=1}^{n}f{\big (}a+k{\tfrac {b-a}{n}}{\big )}\right]$

In some simple cases, this expression can be reduced to a real number, which can be interpreted as the area under the curve if $f(x)$ is positive on $[a,b]$ .

Example 1

Find $\int \limits _{0}^{2}x^{2}dx$ by writing the integral as a limit of Riemann sums.

$\int \limits _{0}^{2}x^{2}dx$	$=\lim _{n\to \infty }\left[{\frac {b-a}{n}}\sum _{k=1}^{n}f(x_{k}^{*})\right]$
	$=\lim _{n\to \infty }\left[{\frac {2}{n}}\sum _{k=1}^{n}f{\big (}{\tfrac {2k}{n}}{\big )}\right]$
	$=\lim _{n\to \infty }\left[{\frac {2}{n}}\sum _{k=1}^{n}\left({\frac {2k}{n}}\right)^{2}\right]$
	$=\lim _{n\to \infty }\left[{\frac {2}{n}}\sum _{k=1}^{n}{\frac {4k^{2}}{n^{2}}}\right]$
	$=\lim _{n\to \infty }\left[{\frac {8}{n^{3}}}\sum _{k=1}^{n}k^{2}\right]$
	$=\lim _{n\to \infty }\left[{\frac {8}{n^{3}}}\cdot {\frac {n(n+1)(2n+1)}{6}}\right]$
	$=\lim _{n\to \infty }\left[{\frac {4}{3}}\cdot {\frac {2n^{2}+3n+1}{n^{2}}}\right]$
	$=\lim _{n\to \infty }\left[{\frac {8}{3}}+{\frac {4}{n}}+{\frac {4}{3n^{2}}}\right]$
	$={\frac {8}{3}}$

In other cases, it is even possible to evaluate indefinite integrals using the formal definition. We can define the indefinite integral as follows:

$\int f(x)dx$	$=\int \limits _{0}^{x}f(t)dt=\lim _{n\to \infty }\left[{\frac {x-0}{n}}\sum _{k=1}^{n}f(t_{k}^{*})\right]$
	$=\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}f{\big (}0+{\tfrac {k(x-0)}{n}}{\big )}\right]$
	$=\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}f{\big (}{\tfrac {kx}{n}}{\big )}\right]$

Example 2

Suppose $f(x)=x^{2}$ , then we can evaluate the indefinite integral as follows.

$\int \limits _{0}^{x}f(t)dt$	$=\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}f{\big (}{\tfrac {kx}{n}}{\big )}\right]$
	$=\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}\left({\frac {kx}{n}}\right)^{2}\right]$
	$=\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}{\frac {k^{2}\cdot x^{2}}{n^{2}}}\right]$
	$=\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\sum _{k=1}^{n}k^{2}\right]$
	$=\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\sum _{k=1}^{n}k^{2}\right]$
	$=\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\cdot {\frac {n(n+1)(2n+1)}{6}}\right]$
	$=\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\cdot {\frac {2n^{3}+3n^{2}+n}{6}}\right]$
	$=x^{3}\cdot \lim _{n\to \infty }\left[{\frac {2n^{3}}{6n^{3}}}+{\frac {3n^{2}}{6n^{3}}}+{\frac {n}{6n^{3}}}\right]$
	$=x^{3}\cdot \lim _{n\to \infty }\left[{\frac {1}{3}}+{\frac {1}{2n}}+{\frac {1}{6n^{2}}}\right]$
	$=x^{3}\cdot \left({\frac {1}{3}}\right)$
	$={\frac {x^{3}}{3}}$

Resources

• Infinite Sums, Wikibooks: Calculus/Integration techniques

• Formal Definition of the Limit, Wikibooks: Calculus

• The Limit of a Function PowerPoint file created by Dr. Sara Shirinkam, UTSA.

• The Limit of a Function notes created by Instructor Beatty, UTSA.

• The Left and Right Hand Limit notes created by Instructor Beatty, UTSA.

• The Limit of a Function Worksheet 1

• The Limit of a Function Worksheet 2

• What is a Limit? Basic Idea of Limits by patrickJMT

• Lots of Limit Examples, Part 1 by patrickJMT

Licensing

Content obtained and/or adapted from:

• Infinite Sums, Wikibooks: Calculus/Integration techniques under a CC BY-SA license
• Formal Definition of the Limit, Wikibooks: Calculus under a CC BY-SA license