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Section Limits and Continuity

Definition 1.14.

If f is a function and each of \(a\) and \(L\) are numbers, then we say that the limit of \(f\) as \(x\) approaches \(a\) is \(L\) if the values for \(f(x)\) get close to the number \(L\) as the values for \(x\) get close to the number \(a.\) We write this as

\begin{equation*} \lim_{x \rightarrow a} f(x) = L \; \; \; \mbox{ or } \; \; \; f(x) \rightarrow L \mbox{ as } x \rightarrow a. \end{equation*}
Example 1.15.

What number is the function \(\dsp f(x) = \frac{x^5-32}{x-2},\) approaching as \(x\) approaches \(2\text{?}\)

x f(x)
\(1.9\) \(72.39\)
\(1.99\) \(79.20\)
\(1.999\) \(79.92\)
\(1.9999\) \(79.992\)
\(2.1\) \(88.41\)
\(2.01\) \(80.80\)
\(2.001\) \(80.08\)
\(2.0001\) \(80.008\)
Table 1.16. First Limit Table

Table 1.16 shows that as \(x\) approaches \(2\) from either the left or the right, \(f(x)\) approaches \(80.\) From this table, we assume that the limit as \(x\) approaches \(2\) of the function \(\displaystyle{f(x) = \frac{x^5-32}{x-2}}\) is \(80.\) Our notation for this will be:

\begin{equation*} \lim_{x\rightarrow 2} \frac{x^5-32}{x-2} = 80. \end{equation*}

This is an intuitive approach to the notion of a limit because we did not give specific meaning to the phrases “x approaches 2” or “f(x) approaches 80.” To give a mathematically accurate definition for the word “limit” took mathematicians almost 200 years, yet our intuitive approach will serve us well for this course. If you are interested in a more precise definition of the limit, read the appendix Limit Definition  or take Real Analysis.

Problem 1.17.

Compute \(\dsp \lim_{x\rightarrow 2} \frac{x^3-8}{x-2}\) by constructing a limit table like Table 1.16. What is \(\dsp \lim_{x\rightarrow -2} \frac{x^3-8}{x-2}\text{?}\)

Problem 1.18.

Arguments for trigonometric functions in calculus are in radians unless stated otherwise, so be sure your calculator is in radian mode. Compute \(\dsp \lim_{x\rightarrow 0} \frac{\sin(x)}{2x}\) by constructing a limit table.

Example 1.19.

Computing limits using algebra.

A reasonable attempt to find the limit of the function \(f\) as \(x\) approaches \(a\) would be to simply plug \(a\) in for \(x\) in the function. As illustrated in the last two problems, this does not always work. There is an important distinction between the value \(f(a)\) and the limit of \(f\) as \(x\) approaches \(a.\) In addition to tables and graphing, we may also use algebra to evaluate some limits.

\begin{equation*} \lim_{x \rightarrow 3} \frac{x^2 - 9}{x - 3} = \lim_{x \rightarrow 3} \frac{(x-3)(x+3)}{x - 3} = \lim_{x \rightarrow 3} \left( x+3 \right) = 6 \end{equation*}
Problem 1.20.

Graph the two functions \(f(x) = x+3\) and \(\dsp g(x) =\frac{x^2-9}{x-3},\) and list their domains. Are \(f\) and \(g\) the same function?

Problem 1.21.

Compute the following limit: \(\dsp \lim_{x \rightarrow 2} \frac{x^2 -4}{x - 2}.\)

Problem 1.22.

Compute these limits and make a note that you will need them in a later problem.

  1. \(\dsp \lim_{t\rightarrow 0}\frac{\sin(t)}{t}.\)

  2. \(\dsp \lim_{t \rightarrow 0} \frac{1 - \cos(t)}{t}\)

Problem 1.23.

Let \(x\) be a real number and compute \(\dsp \lim_{h \rightarrow 0} \frac{hx^2 + h}{h}\text{.}\) Your answer will be a number written in terms of the number \(x.\)

Problem 1.24.

Compute the following limits.

  1. \(\dsp \lim_{x \rightarrow 3} \frac{x^3-27}{x-3}\)

  2. \(\dsp \lim_{x \rightarrow 1} \frac{\frac{1}{x} - 1}{x^2-1}\)

  3. \(\dsp \lim_{x \rightarrow -2} \frac{x^3+3x^2+3x+2}{x^2+x-2}\)

There are rules (theorems) that will facilitate the computation of limits. The next problem will help you find these basic rules.

Problem 1.25.

Read Problem 1.26 and then compute the following limits.

  1. \(\dsp \lim_{x \rightarrow 3} 5\)

  2. \(\dsp \lim_{x \rightarrow 3} x\)

  3. \(\dsp \lim_{x \rightarrow \frac{\pi}{2}} 6\sin(x)\) and \(\dsp 6\lim_{x \rightarrow \frac{\pi}{2}} \sin(x)\)

  4. \(\dsp (\lim_{x \rightarrow 4} x^2 + \lim_{x \rightarrow 4} 4x^5)\) and \(\dsp \lim_{x \rightarrow 4} (x^2 + 4x^5)\)

  5. \(\dsp \left( \lim_{x \rightarrow -3} x^2 \right) \cdot \left( \lim_{x \rightarrow -3} \frac{1}{x} \right)\) and \(\dsp \lim_{x \rightarrow -3} \left( x^2 \cdot \frac{1}{x} \right)\)

  6. \(\dsp\lim_{x \rightarrow 5} \frac{x^2}{x-4}\) and \(\dsp \frac{ \lim_{x\rightarrow 5} x^2 }{ \lim_{x \rightarrow 5} x-4 }\)

  7. \(\dsp\lim_{x \rightarrow 2} \sqrt{x^{2}+x+3}\) and \(\dsp \sqrt{\lim_{x \rightarrow 2} (x^{2}+x+3)}\)

  8. \(\dsp\lim_{x \rightarrow -1} (4x+1)^{3}\) and \(\dsp [\lim_{x \rightarrow -1} (4x+1)]^{3}\)

Problem 1.26.

Write five theorems (facts) about limits from your observations in the previous problems.

Problem 1.27.

Graph and state the domain for the following functions.

  1. \(\dsp p(x) = 3x^2 - 6x\)

  2. \(\dsp f(x) = \left\{ \begin{array}{ll} -2x-4 \amp \mbox{if} \; \; \; \; x \lt 1 \\ x+1 \amp \mbox{if} \; \; \; \; x > 1 \end{array} \right.\)

  3. \(\dsp g(x) = \left\{ \begin{array}{ll} x^2+2 \amp \mbox{if} \; \; \; \; x \lt 1 \\ 0 \amp \mbox{if} \; \; \; \; x = 1 \\ 2x+1 \amp \mbox{if} \; \; \; \; x > 1 \end{array} \right.\)

Sometimes we are interested in the limit of a function \(f\) as \(x\) approaches \(a\) from the left or right side. To make clear which side we are discussing, we write \(\dsp \lim_{x \rightarrow a^-} f(x)\) (left-hand limit) or \(\dsp \lim_{x \rightarrow a^+} f(x)\) (right-hand limit). If both the left- and right-hand limits exist and are equal, then the limit exists. Otherwise, the limit does not exist. This can happen when either the left-hand limit or the right-limit does not exist, or when both exist but are not equal.

Problem 1.29.

Using the functions from the previous problem, compute the following limits if they exist. If they do not exist, state why.

  1. \(\lim_{x \rightarrow -3^+} p(x), \lim_{x \rightarrow -3^-} p(x), \mbox{ and } \lim_{x \rightarrow -3} p(x)\)

  2. \(\lim_{x \rightarrow 1^-} f(x), \lim_{x \rightarrow 1^+} f(x), \mbox{ and } \lim_{x \rightarrow 1} f(x)\)

  3. \(\lim_{x \rightarrow 1^-} g(x), \lim_{x \rightarrow 1^+} g(x), \mbox{ and } \lim_{x \rightarrow 1} g(x)\)

Definition 1.30.

If \(f\) is a function and \(a\) is in the domain of \(f\) then we say that f is continuous at a if \(\lim_{x \rightarrow a} f(x) = f(a)\text{.}\)

There are three subtle points in this definition. The definition requires that for \(f\) to be continuous at \(a,\) each of the following conditions must be met.

  1. \(a\) must be in the domain of \(f\) (otherwise \(f(a)\) doesn't exist),

  2. \(\dsp \lim_{x \rightarrow a} f(x)\) must exist, and

  3. these two numbers must be equal.

The absolute value function is an example of what we call a piecewise defined function.

Definition 1.31.

The absolute value function is defined by:

\begin{equation*} |x| = \left\{ \begin{array}{ll} +x \amp \mbox{if} \;\;\;\; x \geq 0 \\ -x \amp \mbox{if} \;\;\;\;x \lt 0 \end{array} \right. \end{equation*}
Problem 1.32.

Determine whether the following functions are continuous at the indicated points.

  1. \(\dsp f(x) = \frac{1}{x+2} \;\;\;\;\; \mbox{at} \;\;\; x=-2.\)

  2. \(\dsp g(x) = \left\{ \begin{array}{ll} -x-1 \amp \mbox{if} \;\;\;\; x \lt 1 \\ x-3 \amp \mbox{if} \;\;\;\; x \geq 1 \end{array} \right. \;\;\;\; \mbox{at} \;\;\;\; x=1.\)

  3. \(\dsp k(x) = \left\{ \begin{array}{ll} \frac{x^2-4}{x-2} \amp \mbox{if} \;\;\;\;x \neq 2 \\ \\ 5 \amp \mbox{if} \;\;\;\; x = 2 \end{array} \right. \;\;\;\;\;\; \mbox{at} \;\;\;\;\; x=1.\)

  4. \(f(x)=4x^{3}+x^{2}-6x-7\) at \(x=-2\)

  5. \(\dsp C(x)=\cos(x)\) at \(x=\frac{\pi }{3}\)

  6. \(\dsp a(x) = \frac{|x-2|}{x-2} \;\;\;\;\; \mbox{at} \;\;\;\; x=2.\)

Definition 1.33.

We say that a function is continuous on a set (such as an open interval) if it is continuous at every point in the set.

Problem 1.34.

For the following, list the open intervals on which the function is continuous.

  1. \(f(x)=x^{2}+5x+3\)

  2. \(\dsp g(x)=\frac{3x-1}{2x+5}\)

  3. \(H(x)=\sin(x)+5x+\sqrt{1-x}\)

  4. \(T(z)=2+\tan(z)\)

Problem 1.36.

Compute the following limits:

  1. \(\dsp \lim_{h \rightarrow 0} \frac{9 h^2 - 3h}{h}\)

  2. \(\dsp \lim_{h \rightarrow 0} \frac{x^2h^2 - xh}{h},\) where \(x\) is some real number.

  3. \(\dsp \lim_{x \rightarrow 2} \left( 2xt + 3x^2t^2 \right),\) where t is some real number.

Problem 1.37.

Let \(f(x) = x^2 + x + 1.\)

  1. Compute and simplify \(\dsp \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}.\)

  2. Compute and simplify \(\dsp \lim_{t \rightarrow x} \frac{f(t) - f(x)}{t-x}.\)

Example 1.38.

Graph \(f(x) = x^2, g(x) = -x^2,\) and \(\dsp b(x) = x^2\sin \left( \frac{1}{x} \right).\) Compute \(\dsp \lim_{x \rightarrow 0} f(x)\) and \(\dsp \lim_{x \rightarrow 0} g(x).\) What can you conclude about \(\dsp \lim_{x \rightarrow 0} b(x)?\)

Example 1.39.

Show that \(\dsp \lim_{h \rightarrow 0} \frac{\sin(h)}{h} =1\) using the unit circle and starting with the fact that \(\sin(h) \leq h \leq \tan(h)\text{.}\)

The previous two examples illustrate the Squeeze Theorem. The importance of the second example will become clear later in the course when we take the derivative of the trigonometric function, sine. The notation \(f \in C_{[a,b]}\) translates as “\(f\) is an element of the set of all continuous function on the interval \([a,b],\)” which means that \(f\) is continuous at every number \(x\) in the interval \([a,b].\) When we write \(g\) is between \(f\) and \(h,\) on \([a,b]\) we mean that for every number \(x \in [a,b]\) we have either \(f(x) \leq g(x) \leq h(x)\) or \(h(x) \leq g(x) \leq f(x).\)

Problem 1.41.

Use the Squeeze Theorem to solve each of these problems.

  1. Compute \(\dsp \lim_{x\rightarrow 0} \left( x^2\cos(x) \right).\)

  2. Compute \(\dsp \lim_{x \rightarrow 0} x\sin \left(\frac{1}{x} \right)\) using \(\dsp f(x) = -x\text{,}\) \(\dsp g(x) = x\sin \left(\frac{1}{x} \right)\text{,}\) and \(\dsp h(x) = x.\)

  3. Compute \(\dsp \lim_{x \rightarrow \infty} \frac{x}{x^2+x+1}\) using \(\dsp f(x) = \frac{1}{x}\text{,}\) \(\dsp g(x) = \frac{x}{x^2+x+1}\text{,}\) and \(\dsp h(x) = 0.\)

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