The limit concept was not fully clarified until the nineteenth century. The French mathematician Augustin-Louis Cauchy (1789–1857, pronounced Koh-shee) gave the following verbal definition: “When the values successively attributed to the same variable approach a fixed value indefinitely, in such a way as to end up differing from it by as little as one could wish, this last value is called the limit of all the others. So, for example, an irrational number is the limit of the various fractions which provide values that approximate it more and more closely.” (Translated by J. Grabiner)

The first mathematically precise definition is due to the German mathematician Karl Weierstraß (1815-1897), given in his Berlin lectures.

DEFINITION Limit

Assume that \(f(x)\) is defined for all \(x\) in an open interval containing \(c\), but not necessarily at \(c\) itself. We say that

if \(|f(x) − L|\) becomes arbitrarily small when \(x\) is any number sufficiently close (but not equal) to \(c\). In this case, we write \[\lim\limits_{x\rightarrow c} f(x) = L\]

We also say that \(f(x)\) approaches or converges to \(L\) as \(x\rightarrow c\) (and we write \(f(x) \rightarrow L\)).

EXAMPLE 1

Use the definition above to verify the following limits:

(a) \(\displaystyle \lim \limits_{x \rightarrow 7} 5=5\)

(b) \(\displaystyle \lim \limits_{x \rightarrow 4} (3x+1)=13\)

Solution

(a) Let \(f(x)=5\). To show that \(\displaystyle \lim \limits_{x \rightarrow 7} 5=5\), we must show that \(|f(x) - 5|\) becomes arbitrarily small when \(x\) is sufficiently close (but not equal) to 7. But observe that \(|f(x) - 5| = |5-5| = 0\) for all \(x\), so what we are required to show is automatic (and it is not necessary to take \(x\) close to 7).

(b) Let \(f(x) = 3x+1\). To show that \(\displaystyle \lim \limits_{x \rightarrow 4} 3x+1=13\), we must show that \(|f(x) – 13|\) becomes arbitrarily small when \(x\) is sufficiently close (but not equal) to 4. We have \[|f(x) - 13| = |(3x+1) - 13| = |3x - 12| = 3|x-4|\] Because \(|f(x) – 13|\) is a multiple of \(|x– 4|\), we can make \(|f(x) – 13|\) arbitrarily small by taking \(x\) sufficiently close to 4.

THEOREM 1

For any constants \(k\) and \(c\), (a) \(\displaystyle \lim \limits_{x \rightarrow c} k=k\) and (b) \(\displaystyle \lim \limits_{x \rightarrow c} x=c\).

Graphical and numerical investigations provide evidence for a limit, but they do not prove that the limit exists or has a given value. This is done for instance by using the Limit Laws or other theorems established in the following sections.

The undefined expression \(\tfrac{0}{0}\) is referred to as an “indeterminate form.”

EXAMPLE 1

Investigate \(\lim\limits_{x\rightarrow 9}\frac{x-9}{\sqrt{x}-3}\) graphically and numerically.

Solution The function \(f(x) = \frac{x-9}{\sqrt{x}-3}\) is undefined at \(x=9\) because the formula for \(f(9)\) leads to the undefined expression \(\tfrac{0}{0}\). Therefore, the graph in Figure 2.9 has a gap at \(x=9\). However, the graph suggests that \(f(x)\) approaches 6 as \(x\rightarrow 9\).

For numerical evidence, we consider a table of values of \(f(x)\) for \(x\) approaching 9 from both the left and the right. Table 2.4 confirms our impression that \[\lim\limits_{x\rightarrow 9}\frac{x-9}{\sqrt{x}-3} = 6\]

Figure 2.9: Graph of \(f(x) = \frac{x-9}{\sqrt{x}-3}\) .

EXAMPLE 2 Limit Equals Function Value

Investigate \(\lim\limits_{x\rightarrow 4} x^2\).

SolutionFigure 2.10 and Table 2.5 both suggest that \(\lim\limits_{x\rightarrow 4} x^2 = 16\). But \(f(x) = x^2\) is defined at \(x=4\) and \(f(4)=16\), so in this case, the limit is equal to the function value. This pleasant conclusion is valid whenever \(f(x)\) is a continuous function, a concept treated in Section 2.2.5.

Figure 2.10: Graph of \(f(x)=x^2\). The limit is equal to the function value \(f(4) = 16\).

EXAMPLE 3

Reinvestigate \(\lim\limits_{x\rightarrow 9}\dfrac{x-9}{\sqrt{x}-3}\) algebraically.

Solution We will algebraically rewrite the expression using a technique called "rationalizing the denominator" to obtain clear solid evidence as to why the guess for the value of the limit in example one above is correct: \[\lim\limits_{x\rightarrow 9}\frac{x-9}{\sqrt{x}-3} = \lim\limits_{x\rightarrow 9}\frac{x-9}{\sqrt{x}-3} \cdot \frac{\sqrt{x}+3}{\sqrt{x}+3} = \lim\limits_{x\rightarrow 9}\frac{{(x-9)}(\sqrt{x}+3)}{{(x-9)}\cdot 1} = \lim\limits_{x\rightarrow 9} \sqrt{x} + 3\]

As in example two, we can evaluate the last limit by finding the function value of \(g(x) = \sqrt{x} + 3\) when \(x=9\), which evaluates to 6. Note that this was not possible before since this would have required us to divide by zero. The purpose of the algebraic rewrite was to eliminate this problem. Why this is justified in this case will be discussed in the next sections.

EXAMPLE 4

Investigate \(\lim\limits_{x\rightarrow 0} \dfrac{\sin x}{x}\) graphically and numerically.

To investigate this limit, consider the graph of \(f(x) = \frac{\sin x}{x}\) given in Figure 2.11. The graph gives the unmistakable impression that \(f(x)\) gets closer and closer to 1 as \(x\rightarrow 0^+\) and as \(x\rightarrow 0^-\).

This conclusion is supported by the table of values for \(f(x)\) for \(x\) near 0 in Table 2.6. Therefore both the graphical and numerical evidence suggest that \(\lim\limits_{x\rightarrow 0} \frac{\sin x}{x} = 1\).

Figure 2.11: Graph of \(f(x) = \frac{\sin x}{x}\).

CAUTION Numerical investigations are often suggestive, but may be misleading in some cases. If, in Example 5, we had chosen to evaluate \(f(x) = \sin\frac{\pi}{x}\) at the values \(x = 0.1, 0.01, 0.001,\ldots\) we might have concluded incorrectly that \(f(x)\) approaches the limit \(0\) as \(x\rightarrow 0\). The problem is that \(f(10^{-n}) = \sin(10^n\pi) = 0\) for every whole number \(n\), but \(f(x)\) itself does not approach any limit.

EXAMPLE 5 A Limit That Does Not Exist

Investigate \(\lim\limits_{x\rightarrow 0} \sin\frac{\pi}{x}\) graphically and numerically.

Solution The function \(f(x) = \sin\frac{\pi}{x}\) is not defined at \(x=0\), but Figure 2.12 suggests that it oscillates between +1 and −1 infinitely often as \(x\rightarrow 0\). It appears, therefore, that \(\lim\limits_{x\rightarrow 0} \sin\frac{\pi}{x}\) does not exist. This impression is confirmed by Table 2.7, which shows that the values of \(f(x)\) bounce around and do not tend toward any limit \(L\) as \(x\rightarrow 0\).

Figure 2.12: Graph of \(f(x) = \sin\frac{\pi}{x}\).

THEOREM 1

The limit itself exists if and only if both one-sided limits exist and are equal.

EXAMPLE 6 Left- and Right-Hand Limits Not Equal

Investigate the one-sided limits of \(f(x) = \frac{x}{|x|}\) as \(x\rightarrow 0\). Does \(\lim\limits_{x\rightarrow 0} f(x)\) exist?

Solution For \(x < 0\), \[f(x) = \frac{x}{|x|} = \frac{x}{-x} = -1\]

Figure 2.13: Graph of \(f(x) = \frac{x}{|x|}\).

Therefore, the left-hand limit is \(\lim\limits_{x\rightarrow 0^-} f(x) = -1\). But for \(x > 0\), \[f(x) = \frac{x}{|x|} = \frac{x}{x} = 1\]

Therefore, \(\lim\limits_{x\rightarrow 0^+} f(x) = 1\). These one-sided limits are not equal, so \(\lim\limits_{x\rightarrow 0} f(x)\) does not exist.

EXAMPLE 7

The function \(f(x)\) in Figure 2.14 is not defined at \(c = 0, 2, 4\). Investigate the one- and two-sided limits at these points.

Figure 2.14: ---

Solution

• \(c=0\): The left-hand limit \(\lim\limits_{x\rightarrow 0^-} f(x)\) does not seem to exist because \(f(x)\) appears to oscillate infinitely often to the left of \(x=0\). On the other hand, \(\lim\limits_{x\rightarrow 0^+} f(x) = 2\).
• \(c=2\): The one-sided limits exist but are not equal: \(\lim\limits_{x\rightarrow 2^-} f(x) = 3\) and \(\lim\limits_{x\rightarrow 2^+} f(x) = 1\). Therefore, \(\lim\limits_{x\rightarrow 2} f(x)\) does not exist.
• \(c=4\): The one-sided limits exist and both have the value 2. Therefore, the two-sided limit exists and \(\lim\limits_{x\rightarrow 4} f(x) = 2\).

Example 8

The applet below lets you visualize that, as \(x \rightarrow 2\), \(f(x)\) approaches a different value from the left as from the right. Thus, the one-sided limits are not equal, and \(\lim\limits_{x \rightarrow 2} f(x)\) does not exist.

Figure 2.15: An applet to explore a function where the one-sided limits are unequal.

CAUTION To write that \(\lim\limits_{x\rightarrow c}f(x)=\infty\) is really an abuse of notation since in reality \(\lim\limits_{x\rightarrow c}f(x)\) does not exist in this case. However, stating that \(\lim\limits_{x\rightarrow c}f(x) = \infty\) conveys the information that \(f(x)\) becomes larger and larger as \(x\) gets closer and closer to \(c\). This is useful information about the behavior of the function \(y=f(x)\) near \(x=c\) and therefore we allow this abuse of notation.

Example 8

Figure 2.16: Both functions have the \(y\)-axis as vertical asymptote.

EXAMPLE 9

Investigate the one-sided limits graphically:

(a) \(\lim\limits_{x\rightarrow 2^\pm}\dfrac{1}{x-2}\)

(b) \(\lim\limits_{x\rightarrow 0^\pm}\dfrac{1}{x^2}\)

(c) \(\lim\limits_{x\rightarrow 0^+}\ln(x)\)

Solution

(a) The graph in the figure on the left below suggests that \(\lim\limits_{x\rightarrow 2^-}\frac{1}{x-2} = -\infty\) and that \(\lim\limits_{x\rightarrow 2^+}\frac{1}{x-2} = \infty\). Therefore the vertical line \(x=2\) is a vertical asymptote. Why are the one-sided limits different? Because \(F(x)=\frac{1}{x-2}\) is negative for \(x < 2\) (so the limit from the left is \(-\infty\)) and \(f(x)\) is positive for \(x > 2\) (so the limit from the right is \(\infty\)).

(b) The graph in the figure in the middle below suggests that \(\lim\limits_{x\rightarrow 0^\pm}\frac{1}{x^2} = \infty\). Indeed, \(f(x) = \frac{1}{x^2}\) is positive for all \(x\neq 0\) and becomes arbitrarily large as \(x\rightarrow 0\) from either side. The line \(x=0\) is a vertical asymptote.

(c) The graph in the figure on the right below below suggests that \(\lim\limits_{x\rightarrow 0^+} \ln(x)=-\infty\) because \(f(x)=\ln(x)\) is negative for \(0 < x < 1\) and tends to \(-\infty\) as \(x\rightarrow 0^+\). The line \(x=0\) is a vertical asymptote.

Figure 2.17: A graphical investigation of functions with vertical asymptotes.