The Limit Laws

LEMMA

a lemma is an auxiliary theorem, a result that justifies its existence only by virtue of its prominent role in the proof of another theorem.

(1) If

|x-x_0| < \frac{\epsilon}{2} \quad and \quad |y-y_0| < \frac{\epsilon}{2}

then

|(x+y) - (x_0+y_0)| < \epsilon

Proof:

\begin{aligned} \\ |(x+y) - (x_0+y_0)| = |(x-x_0)+ (y-y_0)| \\ \leq |x-x_0| + |y-y_0| \\ < \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ = \epsilon \\ \end{aligned}

(2) If

|x-x_0| < min(1, \frac{\epsilon}{2(|y_0| + 1)}) \quad and \quad |y-y_0| < \frac{\epsilon}{2(|x_0| + 1)}

then

|xy-x_0y_0| < \epsilon

Proof:

Since $|x-x_0| < 1$ we have

|x| - |x_0| \leq |x-x_0| < 1

so that

|x| < 1 + |x_0|

Thus

\begin{aligned} |xy-x_0y_0| = |x(y-y_0) + y_0(x-x_0)| \\ \leq |x| \cdot |y-y_0| + |y_0| \cdot |x-x_0| \\ < (1 + |x_0|) \cdot \frac{\epsilon}{2(|x_0| + 1)} + |y_0| \cdot \frac{\epsilon}{2(|y_0| + 1)} \\ < \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ < \epsilon \\ \end{aligned}

(3) If $y_0 \neq 0$ and

|y-y_0| < min(\frac{|y_0|}{2}, \frac{\epsilon |y_0|^2}{2})

then $y \neq 0$ and

|\frac{1}{y} - \frac{1}{y_0}| < \epsilon

Proof:

We have

|y_0| - |y| \leq |y - y_0| < \frac{|y_0|}{2}

so $|y| > |y_0|/2$ . In particular, $y \neq 0$ , and

\frac{1}{|y|} < \frac{2}{|y_0|}

Thus

|\frac{1}{y} - \frac{1}{y_0}| = \frac{|y_0 - y|}{|y| \cdot |y_0|} < \frac{2}{|y_0} \cdot \frac{1}{|y_0|} \cdot \frac{\epsilon |y_0|^2}{2} < \epsilon

Theorem

\lim_{x \to a} f(x) = l \quad and \quad \lim_{x \to a} g(x) = m

then

\begin{aligned} & (1) \lim_{x \to a} (f+g)(x) = l + m \\ & (2) \lim_{x \to a} (f \cdot g)(x) = l \cdot m \\ & (3) \lim_{x \to a} (\frac{1}{g})(x) = \frac{1}{m} \quad if \quad m \neq 0 \\ \end{aligned}

Proof

The hypothesis means that for every $\epsilon > 0$ there are $\delta_1, \delta_2 > 0$ such that, for all $x$

if \quad 0 < |x-a| < \delta_1, \quad then \quad |f(x) - l| < \epsilon,

and \quad if \quad 0 < |x-a| < \delta_2, \quad then \quad |g(x) - m| < \epsilon,

This means (since, after all, $\epsilon/2$ is also a positive number) that there are $\delta_1, \delta_2 > 0$ such that, for all $x$

if \quad 0 < |x-a| < \delta_1, \quad then \quad |f(x) - l| < \frac{\epsilon}{2},

and \quad if \quad 0 < |x-a| < \delta_2, \quad then \quad |g(x) - m| < \frac{\epsilon}{2},

Now let $\delta = min(\delta_1, \delta_2)$ . if $0 < |x-a| < \delta$ , then $0 < |x-a| < \delta_1$ and $0 < |x-a| < \delta_2$ are both true, so both

|f(x) - l| < \frac{\epsilon}{2} \quad and \quad |g(x) - m| < \frac{\epsilon}{2}

are true. But by part (1) of the lemma this implies that

|(f+g)(x) - (l+m)| < \epsilon

This proves (1).

To prove (2) we proceed similarly, after consulting part (2) of the lemma. If $\epsilon > 0$ there are $\delta_1$ , $\delta_2> 0$ such that, for all $x$ ,

if \quad 0 < |x-a| < \delta_1, \quad and \quad |f(x) - l| < min(1, \frac{\epsilon}{2(|m| + 1)})

and \quad if \quad 0 < |x-a| < \delta_2, \quad and \quad |g(x) - m| < \frac{\epsilon}{2(|l| + 1)}

Again let $\delta = min(\delta_1, \delta_2)$ , if $0 < |x-a| < \delta$ , then

|f(x) - l| < min(1, \frac{\epsilon}{2(|m| + 1)})

|g(x) - m| < \frac{\epsilon}{2(|l| + 1)}

So, by the lemma, $|(f \cdot g)(x) - l \cdot m| < \epsilon$ , and this proves (2)

Finally, if $\epsilon > 0$ there is a $\delta > 0$ such that, for all $x$ ,

if \quad 0 < |x-a| < \delta, \quad then \quad |g(x) - m| < min(\frac{|m|}{2}, \frac{\epsilon |m|^2}{2})

But according to part (3) of the lemma this means, first, that $g(x) \neq 0$ , so $(1/g)(x)$ makes sense, and second that

|(\frac{1}{g})(x) - \frac{1}{m}| < \epsilon

This proves (3).

Reference: Calculus Micheal Spivak. 5. Limits

The Limit Laws

LEMMA

Theorem

Proof

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