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# 13.4: Limit Laws

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When you studied Calculus I your text presented a number of limit laws, such as $$\lim_{x \rightarrow c} f(x)g(x) = (\lim_{x \rightarrow c} f(x)) \cdot (\lim_{x \rightarrow c} g(x))$$. These laws allowed you to compute complex limits by reducing them to simpler limits, until the answer was at hand. But your calculus text probably did not prove the laws. Rather, you were asked to accept them as intuitively plausible (and useful) facts.

Using Definition 13.2, we now present proofs of some limit laws. This serves two purposes. First, it puts your knowledge of calculus on a firmer foundation. Second, it highlights various strategies and thought patterns that are useful in limit proofs, which come to bear in later courses and work.

The inequalities (13.1), (13.2) and (13.3) from page 245 play a crucial role. For convenience we repeat them here. For any $$x, y \in \mathbb{R}$$,

$$|x-y| \le |x|+|y|, |x+y| \le |x|+|y|$$, and $$|x|-|y| \le |x+y|$$.

We will use these frequently, usually without comment. Our first limit law concerns the constant function $$\lim_{x \rightarrow c} f(x) = a$$ where $$a \in \mathbb{R}$$.

Its graph is a horizontal line with y-intercept a. It should be obvious that $$\lim_{x \rightarrow c} f(x) = a$$ for any real number c. Nonetheless, let’s prove this obvious fact.

Theorem 13.2 (Constant function rule)

If $$a \in \mathbb{R}$$, then $$\lim_{x \rightarrow c} a = a$$.

Proof. Suppose $$a \in \mathbb{R}$$. According to Definition 13.2, to prove $$\lim_{x \rightarrow c} a = a$$, we must show that for any $$\epsilon > 0$$, there is a $$\delta > 0$$ for which $$0 < |x-c| < \delta$$ implies $$|a-a| < \epsilon$$. This is almost too easy. Just let $$\delta = 1$$ (or any other number). Then $$|a-a| < \epsilon$$ is automatic, because $$|a-a| = 0$$.

The identity function $$f : \mathbb{R} \rightarrow \mathbb{R}$$ is $$f(x) = x$$. Next we prove $$\lim_{x \rightarrow c} f(x) = c$$.

Theorem 13.3 (Identity function rule)

If $$c \in \mathbb{R}$$, then $$\lim_{x \rightarrow c} x = c$$.

Proof. Given $$\epsilon > 0$$, let $$\delta = \epsilon$$. Then $$0 < |x-c| < \delta$$ implies $$|x-c| < \epsilon$$. By Definition 13.2, this means $$\lim_{x \rightarrow c} x = c$$.

Theorem 13.4 (Constant multiple rule)

If $$\lim_{x \rightarrow c} f(x)$$ exists, and $$a \in \mathbb{R}$$, then $$\lim_{x \rightarrow c} af(x) = a\lim_{x \rightarrow c} f(x)$$.

Proof. Suppose $$\lim_{x \rightarrow c} f(x) = L$$ exists. We must show $$\lim_{x \rightarrow c} af(x) = a\lim_{x \rightarrow c} f(x)$$. If $$a = 0$$, then this reduces to $$\lim_{x \rightarrow c} 0 = 0$$, which is true by Theorem 13.2. Thus, for the remainder of the proof we can assume $$a \ne 0$$.

Suppose $$\lim_{x \rightarrow c} f(x) = L$$. We must prove $$\lim_{x \rightarrow c} af(x) = aL$$. By Definition 13.2, this means we must show that for any $$\epsilon > 0$$, there is a $$\delta > 0$$ for which $$0 < |x-c| < \delta$$ implies $$|af(x)-aL| < \epsilon$$. Let $$\epsilon > 0$$. Because $$\lim_{x \rightarrow c} f(x) = L$$, there exists a $$\delta > 0$$ for which $$0 < |x-c| < \delta$$ implies $$|f(x)-L| < \frac{\epsilon}{|a|}$$. So if $$0 < |x-c| < \delta$$, then $$|af(x)-aL| =a|f(x)-L| = |a| \cdot |f(x)-L| < |a| \frac{\epsilon}{|a|| = \epsilon$$.

In summary, we’ve shown that for any $$\epsilon > 0$$, there is a $$\delta > 0$$ for which $$0 < |x-c| < \delta$$ implies $$|af(x)-aL| < \epsilon$$. By Definition 13.2, $$\lim_{x \rightarrow c} af(x) = aL$$.

Theorem 13.5 (Sum rule)

If both $$\lim_{x \rightarrow c} f(x)$$ and $$\lim_{x \rightarrow c} g(x)$$ exist, then $$\lim_{x \rightarrow c} (f(x)+g(x)) = \lim_{x \rightarrow c} f(x)+\lim_{x \rightarrow c} f(x)$$.

Proof. Let $$\lim_{x \rightarrow c} f(x) = L$$ and $$\lim_{x \rightarrow c} g(x) = M$$. We must prove $$\lim_{x \rightarrow c} (f(x)+g(x)) = L+M$$. To prove this, take $$\epsilon > 0$$. We need to find a corresponding $$\delta$$ for which $$0 < |x-c| < \delta$$ implies $$|(f(x)+g(x))-(L+M)| < \epsilon$$. With this in mind, notice that

$$|(f(x)+g(x))-(L+M)| = |f(x)-L+g(x)-M| \le |f(x)-L|+|g(x)-M|$$. (A)

As $$\lim_{x \rightarrow c} f(x) = L$$, there is a $$\delta' > 0$$ such that $$0 < |x-c| < \delta'$$ implies $$|f(x)-L| <\frac{2}{\epsilon}$$.

As $$\lim_{x \rightarrow c} g(x) = M$$, there is a $$\delta'' > 0$$ such that $$0 < |x-c| < \delta''$$ implies $$|g(x)-M| <\frac{2}{\epsilon}$$.

Now put $$\delta = min \{\delta', \delta''\}$$, meaning that $$\delta$$ equals the smaller of $$\delta'$$ and $$\delta''$$.

If $$0 < |x-c| < \delta$$, then (A) gives $$|f(x)+g(x)-(L+M)| \le \frac{2}{\epsilon}+\frac{2}{\epsilon} = \epsilon$$.

We’ve now shown that for any $$\epsilon > 0$$, there is a $$\delta > 0$$ for which $$0 < |x-c| < \delta$$

implies $$|(f(x)+g(x))-(L+M)| < \epsilon$$. Thus $$\lim_{x \rightarrow c} (f(x)+g(x)) = L+M$$.

Theorem 13.6 (Difference rule)

If both $$\lim_{x \rightarrow c} f(x)$$ and $$\lim_{x \rightarrow c} g(x)$$ exist, then $$\lim_{x \rightarrow c} (f(x)-g(x)) = \lim_{x \rightarrow c} f(x)-\lim_{x \rightarrow c} f(x)$$.

Proof. Combining the sum rule with the constant multiple rule gives

$$\lim_{x \rightarrow c} (f(x)-g(x)) = \lim_{x \rightarrow c} (f(x)+(-1)g(x)) = \lim_{x \rightarrow c} f(x)+\lim_{x \rightarrow c} (-1)g(x)$$

$$= \lim_{x \rightarrow c} f(x)+(-1)\lim_{x \rightarrow c} g(x) = \lim_{x \rightarrow c} f(x)-\lim_{x \rightarrow c} f(x)$$.

Theorem 13.7 (Multiplication rule)

If both $$\lim_{x \rightarrow c} f(x)$$ and $$\lim_{x \rightarrow c} g(x)$$ exist, then $$\lim_{x \rightarrow c} (f(x)g(x)) = (\lim_{x \rightarrow c} f(x)) \cdot (\lim_{x \rightarrow c} f(x))$$.

Proof. Let $$\lim_{x \rightarrow c} f(x) = L$$ and $$\lim_{x \rightarrow c} g(x) = M$$. We must prove $$\lim_{x \rightarrow c} (f(x)g(x)) = LM$$. To prove this, take $$\epsilon > 0$$. We need to find a corresponding $$\delta$$ for which $$0 < |x-c| < \delta$$ implies $$|(f(x)g(x))-(LM)| < \epsilon$$. With this in mind, notice that

$$|(f(x)g(x))-(LM)| = |(f(x)g(x)-Lg(x))+(Lg(x)-LM)| \le |(f(x)g(x)-Lg(x))|+|(Lg(x)-LM)|$$

$$= |(f(x)-L)g(x)|+|L(g(x)-M)|$$

$$= |f(x)-L| \cdot g(x)+|L| \cot |g(x)-M|$$ (A)

Because $$\lim_{x \rightarrow c} f(x) = L$$ and $$\lim_{x \rightarrow c} g(x) = M$$, we can make the expressions $$|f(x)-L|$$ and $$|L| \cot |g(x)-M|$$ in (A) arbitrarily small by making $$|x-c|$$ sufficiently small.

But the term $$|f(x)-L| \cot |g(x)|$$ is a problem. For all we know, $$g(x)$$ could grow large as $$|f(x)-L|$$ shrinks. To deal with this, choose some $$\delta' > 0$$ small enough so that $$0 < |x-c| < \delta'$$ implies $$|g(x)-M| < 1$$. Then as long as $$0 < |x-c| < \delta'$$,

$$g(x) = |(g(x)-M)+M| \le |g(x)-M|+|M| < 1+|M|$$.

Replacing the factor of $$g(x)$$ in (A) with the larger quantity $$1+|M|$$, we get

$$|f(x)g(x)-LM| < |f(x)-L| \cdot (1+|M|)+|L| \cot |g(x)-M|$$, (B)

which holds provided $$0 < |x-c| < \delta'$$.

Choose $$\delta'' > 0$$ such that $$0 < |x-c| < \delta''$$ implies $$|f(x)-L| < \frac{\epsilon}{2(1+|M|)}$$.

Also choose $$\delta''' > 0$$ such that $$0 < |x−c| < \delta'''$$ implies $$|g(x)-M| < \frac{\epsilon}{2(1+|L|)}$$.

Now put $$\delta = min \{\delta', \delta'', \delta'''\}$$. If $$0 < |x-c|< \delta$$, then (B) becomes

$$|f(x)g(x)-LM| < \frac{\epsilon}{2(1+|M|)} \cdot (1+|M|)+|L| \cot \frac{\epsilon}{2(1+|L|)} \le \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$$,

To summarize, we’ve shown that for any $$\epsilon > 0$$, there is a $$\delta > 0$$ for which $$0 < |x-c| < \delta$$ implies $$|f(x)g(x)-LM| < \epsilon$$. Therefore $$\lim_{x \rightarrow c} (f(x)g(x)) = LM$$.

Our final rule has proof similar to that of the multiplication rule. We just have to take a little extra care with the denominators.

Theorem 13.8 (Division rule)

If both $$\lim_{x \rightarrow c} f(x)$$ and $$\lim_{x \rightarrow c} g(x)$$ exist, and $$\lim_{x \rightarrow c} g(x) \ne 0$$, then $$\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow c} f(x)}{\lim_{x \rightarrow c} f(x)}$$.

Proof. Let $$\lim_{x \rightarrow c} f(x) = L$$ and $$\lim_{x \rightarrow c} g(x) = M$$. We must prove $$\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \frac{L}{M}$$. To prove this, take $$\epsilon > 0$$. We need to find a corresponding $$\delta$$ for which $$0 < |x-c| < \delta$$ implies $$|\frac{f(x)}{g(x)}-\frac{L}{M}| < \epsilon$$. With this in mind, notice that

$$|\frac{f(x)}{g(x)}-\frac{L}{M}| = |\frac{Mf(x)-Lg(x)}{Mg(x)}| = |\frac{(Mf(x)-LM)-(Lg(x)-LM)}{Mg(x)}|$$

$$= |\frac{(f(x)-L)}{g(x)}-\frac{L}{Mg(x)}(g(x)-M)|$$

$$\le |\frac{(f(x)-L)}{g(x)}|+|\frac{L}{Mg(x)}(g(x)-M)|$$

$$\le \frac{1}{|g(x)|} \cdot |f(x)-L| + \frac{1}{|g(x)|} \cdot \frac{|L|}{|M|} \cdot |g(x)-M|$$ (A)

Because $$\lim_{x \rightarrow c} f(x) = L$$ and $$\lim_{x \rightarrow c} g(x) = M$$, we can make the expressions $$|f(x)-L|$$ and $$\frac{|L|}{|M|} \cdot |g(x)-M|$$ in (A) arbitrarily small by making $$|x-c|$$ sufficiently small. To deal with the factor $$\frac{1}{|g(x)|}$$, choose $$\delta' > 0$$ so that $$0 < |x-c| < \delta'$$ implies

$$|g(x)-M| < \frac{|M|}{2}$$. So if $$0 < |x-c| < \delta'$$, the inequality (13.3) assures us

$$g(x) = |(g(x)-M)+M| \ge |M|-|g(x)-M| > |M|-\frac{|M|}{2} = \frac{|M|}{2}$$.

That is, $$g(x) > \frac{|M|}{2}$$, and consequently $$\frac{1}{|g(x)|} < \frac{2}{|M|}$$. Replacing the occurrences of $$\frac{1}{|g(x)|}$$ in (A) with the larger value

$$\frac{2}{|M|}$$, we get

$$|\frac{f(x)}{g(x)}-\frac{L}{M}| < \frac{2}{|M|} \cdot |f(x)-L|+|\frac{2L}{M^2}| \cdot |g(x)-M|$$, (B)

which holds provided $$0 < |x-c| < \delta'\. Two cases finish the proof. Case1. Suppose \(L \ne 0$$. Choose $$\delta'' > 0$$ so $$0 < |x-c| < \delta''$$ implies $$|f(x)-L| < \frac{\epsilon|M|}{4}$$.

Also choose $$\delta''' > 0$$ so that $$0 < |x-c| < \delta'''$$ implies $$|g(x)-M| < \epsilon |\frac{M^2}{4L}|$$. Put $$\delta = min \{\delta', \delta'', \delta'''\}$$. If $$0 < |x-c| < \delta$$, then (B) yields

$$|\frac{f(x)}{g(x)}-\frac{L}{M}| < \frac{2}{|M|} \cdot \frac{\epsilon|M|}{4}+|\frac{2L}{M^2}| \cdot \epsilon\frac{M^2}{4L} = \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$$

Case 2. Suppose $$L = 0$$. Let $$\delta'' > 0$$ be such that $$0 < |x-c| < \delta''$$ implies $$|f(x)-L| < \frac{\epsilon|M|}{2}$$. Putting $$\delta = min \{\delta', \delta''\}$$, the inequality (B) becomes

$$|\frac{f(x)}{g(x)}-\frac{L}{M}| < \frac{2}{|M|} \frac{\epsilon|M|}{2} = \epsilon$$

In each case we have shown that for any $$\epsilon > 0$$, there is a $$\delta > 0$$ for which $$0 < |x-c| < \delta$$ implies $$|\frac{f(x)}{g(x)}-\frac{L}{M}| < \epsilon$$, so the proof is finished.

Though you may not have proved any limit laws in your calculus course, you used them extensively. A common situation involved $$\lim_{x \rightarrow c} f(x)$$, where $$f(c)$$ was undefined because of a zero denominator. You learned to overcome this by algebraically canceling the offending part of the denominator.

Example 13.6

Find $$\lim_{x \rightarrow 1} \frac{\frac{1}{x}-1}{1-x}$$.

Here x approaches 1, but simply plugging in $$x = 1$$ gives $$\frac{\frac{1}{1}-1}{1-1} = \frac{0}{0}$$ (undefined). So we apply whatever algebra is needed to cancel the denominator $$1-x$$, and follow this with limit laws:

$$\lim_{x \rightarrow 1} \frac{\frac{1}{x}-1}{1-x} = \lim_{x \rightarrow 1} \frac{\frac{1}{x}-1}{1-x} \cdot \frac{x}{x}$$.

$$= \lim_{x \rightarrow 1} \frac{(1-x)}{(1-x)x}$$

$$= \lim_{x \rightarrow 1} \frac{1}{x}$$

$$= \frac{\lim_{x \rightarrow 1} 1}{\lim_{x \rightarrow 1} x} = \frac{1}{1} = 1$$.

## Exercise

Exercise $$\PageIndex{1}$$

Given two or more functions $$f_{1}, f_{2}, \cdots, f_{n}$$, suppose that $$\lim_{x \rightarrow c} f_{i}(x)$$ exists for each $$1 \le i \le n$$.

Prove that $$\lim_{x \rightarrow c} f_{1}(x)+f_{2}(x)+\cdots+f_{n}(x) = \lim_{x \rightarrow c} f_{1}(x)+\lim_{x \rightarrow c} f_{2}(x)+\cdots+\lim_{x \rightarrow c} f_{n}(x)$$. Use induction on n, with Theorem 13.5 serving as the base case.

Exercise $$\PageIndex{2}$$

Given two or more functions $$f_{1}, f_{2}, \cdots, f_{n}$$, suppose that $$\lim_{x \rightarrow c} f_{i}(x)$$ exists for each $$1 \le i \le n$$.

Prove that $$\lim_{x \rightarrow c} (f_{1}(x)f_{2}(x) \cdots f_{n}(x)) = (\lim_{x \rightarrow c} f_{1}(x)) \cdot (\lim_{x \rightarrow c} f_{2}(x)) \cdots (\lim_{x \rightarrow c} f_{n}(x))$$. Use induction on n, with Theorem 13.7 serving as the base case.

Exercise $$\PageIndex{3}$$

Use the previous two exercises and the constant multiple rule (Theorem 13.4) to prove that that if $$f(x)$$ is a polynomial, then $$\lim_{x \rightarrow c} f(x) = f(c)$$ for any $$c \in \mathbb{R}$$.

Exercise $$\PageIndex{4}$$

Use Exercise 3 with a limit law to prove that if $$\frac{f(x)}{g(x)}$$ is a rational function (a polynomial divided by a polynomial), and $$g(c) \ne 0$$, then $$\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \frac{f(c)}{g(c)}$$

Exercise $$\PageIndex{5}$$

Use Definition 13.2 to prove that limits are unique in the sense that if $$\lim_{x \rightarrow c} f(x) = L$$ and $$\lim_{x \rightarrow c} f(x) = M$$, then $$L = M$$.

Exercise $$\PageIndex{6}$$

Prove the squeeze theorem: Suppose $$g(x) \le f (x) \le h(x)$$ for all $$x \in \mathbb{R}$$ satisfying $$0 < |x-c| < \delta$$ for some $$\delta > 0$$. If $$\lim_{x \rightarrow c} g(x) = L = \lim_{x \rightarrow c} h(x)$$, then $$\lim_{x \rightarrow c} f(x) = L$$.