# Limits (Calculus Starts with Limits)

• By David A. Smith, Founder & CEO, Direct Knowledge
• David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

Limits are used to study the behavior of quantities under a process of change. For example, limits can be used to describe the behavior of a function on its domain. Here we study one-sided limits and two-sided limits with emphasis on graphs. We discuss unbounded behavior and oscillating behavior with many examples given. Hi

## An Intuitive Introduction to Limits

We illustrate when a limit does not exist by giving three case examples.

• A limit does not exist because the one-sided limits do not agree in value.
• A limit does not exist because of an oscillating behavior of a function, and
• a limit does not exist (no finite value) because of an unbounded behavior of a function.

## Limits Using Tables

A limit is used to describe the behavior of a function near a point but not at the point. The function need not even be defined at the point. If it is defined there, the value of the function at the point does not affect the limit. Intuitively, $$\lim_{x\to c}f(x)=L$$ means we can make$f(x)$ as close to $L$ as we wish by taking any $x$ sufficiently close to, but different from $c.$

Example. Find the limit of $f(x)$ as $x$ approaches $c$ using a table of functional values for $$f(x)=\frac{4x-9}{x^2-4}$$ and $c=3.$

Solution. We compute, $$\begin{array}{ll} \begin{array}{l|l} x & f(x) \\ \hline 2.997 & 0.599758 \\ 2.998 & 0.599839 \\ 2.999 & 0.59992 \end{array} & \begin{array}{l|l} x & f(x) \\ \hline 3.003 & 0.600238 \\ 3.002 & 0.600159 \\ 3.001 & 0.600079 \end{array} \end{array}$$ Thus as $x$ approaches $3$ from the left we estimate that $f(x)$ approaches $3/5$; and as $x$ approaches $3$ from the right we estimate that $f(x)$ approaches $3/5.$ Therefore, $$\lim_{x\to 3}\frac{4x-9}{x^2-4}=\frac{3}{5}.$$ as an estimation.

Example. We will use a guessing method to show why the formal definition of a limit is a necessity. Use a table to guess the values of $$L=\lim_{x\to 0}\frac{2\sqrt{x+1}-x-2}{x^2}.$$

Solution. From the table $$\begin{array}{c|ccccccc} x & -0.5 & -0.1 & -0.01 & -0.001 & 0 & 0.001 & 0.005 \\ \hline f(x) & -0.3431 & -0.2633 & -0.2513 & -0.2501 & \text{-} & -0.2499 & -0.2494 \end{array}$$ The number $L$ is suggested to be $-0.25.$ Interestingly, if you try $$x=0.0000001$$ just to make sure you have taken numbers close enough to 0, you may find that the calculator gives the value 0. Does this mean that the limit is 0? No, the calculator may give you a false answer because when $x$ is small enough (i.e. $0.0000001$ ) then $2\sqrt{x+1}-x-2$ seems like 0. But in fact $f(0.0000001)$ is not equal to 0.

The point is, using technology to verify a computation can lead to misunderstanding; and in fact, a formal definition of a limit is needed. Using the formal definition of a limit, we can prove what the value of the limit is without any doubt. This type of proof is usually called an \emph{epsilon-delta} proof since the formal definition is usually stated with the greek letters $\epsilon$ (epsilon) and $\delta$ (delta).

Example. We will use a guessing method to show why the formal definition of a limit is a necessity. Use tables of values to find the limit $$\lim_{x\to 0}\left(x^3+\frac{\cos 5x}{10,000}\right).$$

Solution. We construct a table of values. $$\begin{array}{c|ccccc} x & 1 & 0.5 & 0.1 & 0.05 & 0.01 \\ \hline f(x) & 1.000028 & 0.124920 & 0.001088 & 0.000222 & 0.000101 \end{array}$$ From the table it appears that $$\lim_{x\to 0}\left(x^3+\frac{\cos 5x}{10,000}\right)=0.$$ However, if we persevere with smaller values of $x,$ the next table suggests $$\lim_{x\to 0}\left(x^3+\frac{\cos 5x}{10,000}\right) =0.0001000=\frac{1}{10,0000}$$ $$\begin{array}{ccc} x & 0.005 & 0.001 \\ \hline f(x) & 1.00010009 & 0.00010000 \end{array}$$ In fact, $$\lim_{x\to 0}\left(x^3+\frac{\cos 5x}{10,000}\right)=\frac{1}{10,0000}$$ which is easily proven once the formal limit definition is used to prove some interesting limit rules and continuity is discussed. In summary, a three-pronged approach to solving limits is often:

## One-Sided and Two-Sided Limits

Consider, for example, a piecewise function with a jump where the function is pieced together (defined or not). Even though the one-sided limits might exist, they must agree in value for the two-sided limit to exist. In short, if the one-sided limits do not agree then the two-sided limit does not exist.

Theorem. The two-sided limit $\lim_{x\to c}f(x)$ exists if and only if the one-sided limits $\lim_{x\to c^+}f(x)$ and $\lim_{x\to c^-}f(x)$ both exist and $$\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x).$$ In which case, $$\lim_{x\to c}f(x)=\lim_{x\to c^+}f(x).$$

Example. Sketch the graph of the piecewise function $f$ defined by $$f(x)= \begin{cases} 2 x^2+1 & x<2 \\ 4 & x=2 \\ 3 x & x>2. \end{cases}$$ Evaluate the following limits, \begin{array}{ccc} \displaystyle\lim_{x\to 2^-}f(x) & \displaystyle\lim_{x\to 2^+}f(x) & \displaystyle\lim_{x\to 2}f(x) \\ \displaystyle\lim_{x\to 1^-}f(x) & \displaystyle\lim_{x\to 1^+}f(x) & \displaystyle\lim_{x\to 1}f(x) \\ \displaystyle\lim_{x\to 4^-}f(x) & \displaystyle\lim_{x\to 4^+}f(x) & \displaystyle\lim_{x\to 4}f(x) \end{array}

Example. Sketch the graph of the piecewise function $g(x)$ where $g$ is defined by $$g(x)= \begin{cases} 3x-2 & x<-1 \\ 4 & x=-1 \\ x+5 & -1 < x < 3 \\ 4 & x=3 \\ 2-x & x > 3 \end{cases}$$ Evaluate the following limits, \begin{array}{ccc} \displaystyle\lim_{x\to -1^-}g(x) & \displaystyle\lim_{x\to -1^+}g(x) & \displaystyle\lim_{x\to -1}g(x) \\ \displaystyle\lim_{x\to 3^-}g(x) & \displaystyle\lim_{x\to 3^+}g(x) & \displaystyle\lim_{x\to 3}g(x) \\ \displaystyle\lim_{x\to 4}g(x) & \displaystyle\lim_{x\to 0}g(x) & \displaystyle\lim_{x\to 1}g(x) \end{array}

## Oscillating Behavior

Also, consider another case where a function has an oscillating behavior. On one hand, the trigonometric functions all have an oscillating (periodic) behavior. However, imagine a function where the oscillation becomes much more pronounced as the variable approaches a fixed point; this type of oscillating behavior is where the function may not have a limit.

Example. Find $\displaystyle\lim_{x\rightarrow 0} \cos \left(\frac{1}{x}\right).$

Solution. The limit does not exist because $\cos (1/x) =1$ and $\cos (1/x) =-1$
for $$x=\frac{1}{2\pi },\frac{1}{4\pi },\frac{1}{6\pi },\ldots \quad \text{and} \quad x=\frac{1}{\pi },\frac{1}{3\pi },\frac{1}{5\pi },\ldots$$ respectively. The graph of $y=\cos (1/x)$ is oscillating around $x=0,$ (see \cref{cosonex}) so we infer that the limit does not exist because$f(x)=\cos (1/x)$ does not approach a number, but rather oscillates, as $x$ approaches 0.

## Unbounded Behavior

Finally, we illustrate the case where a function becomes unbounded as the variable approaches a fixed point; for example, a function with a vertical asymptote. Without a finite number to assign the limit, we sometimes say that the limit does not exist.

Example. Determine $\displaystyle \lim_{x\to 0}\frac{1}{x}.$

Solution. Since $f$ decreases without bound as $x\to 0^-$ and $f$ increases without bound as $x\to 0^+,$ we say that $\lim_{x\to0}f(x)$ does not exist.

Example. Determine $\displaystyle \lim_{x\to 0}\frac{1}{x^2}.$

Solution. As $x\to 0^-$ we see that $f$ increases with $f(x)\to +\infty$ and as $x\to 0^-$ we see that $f$ increases with $f(x)\to +\infty$. Therefore, we say that $\lim_{x\to0}f(x)=+\infty .$

## Exercises on Limits

Exercise. Estimate the limits, if they exist, by using a table of values to two decimal places.

$(1) \quad \displaystyle\lim_{x\to 2^+ }\frac{x^2-4}{x-4}$

$(2) \quad \displaystyle\lim_{x\to 4^+ }\frac{\frac{1}{\sqrt{x}}-\frac{1}{2}}{x-4}$

$(3) \quad \displaystyle\lim_{x\to 0 }\frac{\sin 2x}{x}$

$(4) \quad \displaystyle\lim_{x\to 0}|x|\sin \frac{1}{x}$

Exercise. Sketch the graph of $f$ and $g.$ Then identify the values of $c$ for which $\lim_{x\to c}f(x)$ and $\lim_{x\to c}g(x)$ exist given $$f(x)= \begin{cases} x^2 & x\leq 2 \\ 8-2x & 2 < x < 4 \\ 4 & x \geq 4 \end{cases}$$ $$g(x)= \begin{cases} \sin x & x < 0 \\ 1-\cos x & 0\leq x\leq \pi \\ \cos x & x > \pi \end{cases}$$

Exercise. Find $a$ so that the function $$f(x)= \begin{cases} a x+3 & x\leq 2 \\ 3-x & x>2 \end{cases}$$ satisfies $\lim_{x\to 2}f(x)=1.$

Exercise. Estimate the limits by using tables of values for

$(1) \quad \displaystyle\lim_{x\to 13}\frac{x^3-9x^2-45x-91}{x-13}$

$(2) \quad \displaystyle\lim_{x\to 13}\frac{x^3-9x^2-39x-86}{x-13}$

Then using long division (or synthetic division if you know it) explain why one of the limits exists and the other does not.

Exercise. Consider the function $$f(x) =\frac{|x+1|-|x-1|}{x}.$$ Estimate $\displaystyle \lim_{x\to 0}f(x)$ by evaluating $f$ at $x$-values near 0. Sketch the graph of the function $f.$

Exercise. Explain why $\displaystyle \lim_{x\to 2}\frac{|x-2|}{x-2}$ does not exist.

Exercise. Evaluate the function $$f(x)=x^2-\frac{2^x}{1000}$$ for $x=1,$ $0.8,$ $0.6,$ $0.4,$ $0.2,$ $0.1,$ and $0.05.$ Guess the value of $\lim_{x\to 0}f(x).$ Evaluate the function $$f(x)=x^2-\frac{2^x}{1000}$$ for $x=0.04,$ $0.02,$ $0.01,$ $0.005,$ $0.003,$ and $0.001.$ Guess again.

Exercise. The tabular approach is a convenient device for discussing limits informally, but if it is not used carefully, it can be misleading. For example, for $x>0,$ let $$f(x)=\sin \left(\frac{\pi }{\sqrt{x}}\right)$$ Construct a table showing the value of $x$ and $f(x)$ for $x=4,$ $4/25,$ $4/81,$ $4/169,$ and $4/289.$ Based on this table what would you say about $\lim_{x\to 0^+}f(x)?$ Construct a table showing the value of $x$ and $f(x)$ for $x=4,$ $4/49,$ $4/121,$ $4/225,$ $4/361.$ Based on this table what would you say about $\lim_{x\to 0^+}f(x)?$ Based on your results in (a) and (b) what do you conclude about $\lim_{x\to 0^+}f(x)?$

Exercise. Sketch the graph of the function $$f(x)= \begin{cases} -(x+1)^2+1 & -1\leq x\leq 0 \\ x^2 & 0 < x < 1 \\ 1 & 1 < x< 2 \\ 2 & x=2 \\ 1 & 2 < x \leq 3. \end{cases}$$ and then use the graph to determine which the following statements about the function $y=f(x)$ are true and which are false?

$(1) \quad \displaystyle \lim_{\to -1^+}f(x)=1$

$(2) \quad \displaystyle \lim_{x\to 2}f(x)$ does not exist

$(3) \quad \displaystyle \lim_{x\to 2}f(x)=2$

$(4) \quad \displaystyle \lim_{x\to 1^-}f(x)=2$

$(5) \quad \displaystyle \lim_{x\to 1^+}f(x)=1$

$(6) \quad \displaystyle \lim_{x\to 1}f(x)$ does not exist

$(7) \quad \displaystyle \lim_{x\to 0^+}f(x)=\lim_{x\to 0^-}f(x)$

$(8) \quad \displaystyle \lim_{x\to c}f(x)$ exists at every $c$ in the open interval $(-1,1).$

$(9) \quad \displaystyle \lim_{x\to c}f(x)$ exists at every $c$ in the open interval $(1,3).$

$(10) \quad \displaystyle \lim_{x\to -1^-}f(x)=0$

$(11) \quad \displaystyle \lim_{x\to 3^+}f(x)$ does not exist

Exercise. Sketch the graph of the function $$f(x)= \begin{cases} 3-x & x<2 \\ 2 & x=2 \\ \frac{x}{2} & x>2. \end{cases}$$ and then use the graph to determine the following? Find $\lim_{x\to 2^+}f(x),$ $\lim_{x\to 2^-}f(x),$ and $f(2).$ Does $\lim_{x\to 2}f(x)$ exist? If so, what is it? If not, why not? Find $\lim_{x\to -1^-}f(x)$ and $\lim_{x\to -1^+}f(x).$ Does $\lim_{x\to -1}f(x)$ exist? If so, what is it? If not, why not?

Exercise. Let $\displaystyle g(x)=\sqrt{x}\sin \left(\frac{1}{x}\right).$ Use the graph of $g$ to determine the following. Does $\lim_{x\to 0^+}g(x)$ exist? If so, what is it? If not, why not? Does $\lim_{x\to 0^-}g(x)$ exist? If so, what is it? If not, why not? Does $\lim_{x\to 0}g(x)$ exist? If so, what is it? If not, why not?

Exercise. Sketch the graph of $$f(x)= \begin{cases} x^3 & x\neq 1 \\ 0 & x=1. \end{cases}$$ Find $\displaystyle \lim_{x\to 1^-}f(x)$ and $\displaystyle \lim_{x\to 1^+}f(x).$ Does $\displaystyle \lim_{x\to 1}f(x)$ exist? If so, what is it? If not, why not?

Exercise. Sketch the graph of $$f(x)=\begin{cases}1-x^2 & x\neq 1 \\ 2 & x=1. \end{cases}$$ Find $\displaystyle \lim_{x\to 1^-}f(x)$ and $\displaystyle \lim_{x\to 1^+}f(x).$ Does $\displaystyle \lim_{x\to 1}f(x)$ exist? If so, what is it? If not, why not?

Exercise. Sketch the graph of $$f(x)=\begin{cases}x & -1 \leq x < 0 \quad \text{or} \quad 0 < x \leq 1 \\ 1 & x=0 \\ 0 & x < -1 \quad \text{or} \quad x>1. \end{cases}$$ What is the domain and range of $f?$ At what points $c,$ if any does $\lim_{x\to c}f(x)$ exist? At what points does only the left-hand limit exist? At what points does only the right-hand limit exist?