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I'm having difficulty writing an $\epsilon - \delta$ proof for the following limit:

$\lim_{x\to 4} \frac{x^2-16}{x+\sin x} = 0$

I've factored it to $\frac{(x+4)(x-4)}{x+\sin x} = 0$

and guessed that I need $\delta = \frac{2}{5}\epsilon$ for $|x-4| < \delta \implies |\frac{x^2-16}{x+\sin x}| < \epsilon$

I've also bounded $|x+4|$ by $\delta + 8$ but I don't know how to control $|x + \sin x|$.

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1 Answer 1

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$$|x-4| < \delta$$

$$4-\delta < x < 4+ \delta$$

$$3 - \delta< x+ \sin x < 5 + \delta$$

$$\frac{1}{5+\delta} < \frac{1}{x+\sin x} < \frac{1}{3-\delta}$$

If $\delta < 1$, $-\delta > -1$, $3-\delta > $2, $\frac{1}{3-\delta} < \frac12$

$$\left| \frac{x^2-16}{x+\sin x}\right| \leq \frac12 |x^2-16|$$

Also, if $\delta < 1$, $|x+4|<9,$

$$\frac12 |x^2-16|\leq \frac 92 |x-4|$$

Hence, for example, I can choose

$$\delta = \min ( \frac12, \frac29 \epsilon)$$

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