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I did this using L'Hospital's rule but I'm trying to figure out a way to do it without L'Hospital's just using the fact that $\lim_{x \rightarrow 0} \frac{\sin x}{x}=1$

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    $\begingroup$ $$\frac{1-\frac{\sin x}x}{1+\frac{\sin x}x}\to\frac{1-1}{1+1}=0$$ $\endgroup$ – Did Oct 16 '16 at 20:46
  • $\begingroup$ Using taylor gets you $\sin(x)=x-\frac{x^3}{6}+\mathcal O(x^5)$. This approach works for most limits you'll see! $\endgroup$ – question Oct 16 '16 at 21:02
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You can just divide the numerator and denominator by $x$ to get

$$\lim\limits_{x\rightarrow 0}\,\frac{1-\frac{\sin x}{x}}{1+\frac{\sin x}{x}}$$

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  • $\begingroup$ You should use double dollar signs to make stuff render in the center. Single is inline. Great answer though! $\endgroup$ – Cameron Williams Oct 16 '16 at 20:50
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We have

$sin(x)=x+x\epsilon(x)$ with

$\lim_{x\to 0}\epsilon(x)=0$

thus the function becomes

$$\frac{-x\epsilon(x)}{2x+x\epsilon(x)}$$

$$=\frac{\epsilon(x)}{2+\epsilon(x)}$$

the limit we look for is $0$.

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