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Determine a so that: $\lim_{x\to0} \frac{\tan(ax)}{\sin(x)} = 2$

So far, I have used the L'hopital rule:

$\frac{\frac{1}{a \cos(x)}}{\cos(x)} = \frac{1}{a \cos^3(x)} = 2$

But I am not sure if this is the right way of solving this limit. Can anyone help me?

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    $\begingroup$ Are you sure the limit is to $\infty$, not to $0$? $\endgroup$ – Toby Mak Apr 11 at 7:21
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    $\begingroup$ @TobyMak you're right, my bad! $\endgroup$ – Julius Apr 11 at 7:53
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Assuming, as Toby Mark suggests, that $x$ tends to $0$, it should be $a=2$. One way to show it is indeed l'Hôpital's rule: $$\lim_{x\to0}\frac{\tan(ax)}{\sin(x)}=\frac{\tan0}{\sin0}=\frac{0}{0}\stackrel{\text{l'H}}{=}\lim_{x\to0}\frac{\frac{a}{\cos^2(ax)}}{\cos(x)}=\lim_{x\to0}\frac{a}{\cos^2(ax)\cos(x)}=\frac{a}{\cos^20\cos0}=a.$$

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If yor limit is for $x\rightarrow 0$ then from the Mclaurin expansion (this is the reason why it is important that the limit is for $x$ that goes to $0$, since for $x$ that goes to $\infty$ you can't apply this result) of the two functions involved in the limit we have: $$\tan{ax}=ax+\frac{(ax)^3}{3}+o(x^3)$$ $$\sin{x}=x-\frac{x^3}{6}+o(x^3)$$ So $$\lim_{x\to 0}\frac{\tan{ax}}{\sin{x}}=\lim_{x\to 0} \frac{ax+\frac{(ax)^3}{3}+o(x^3)}{x-\frac{x^3}{6}+o(x^3)}=\lim_{x\to 0}\frac{ax}{x}=a=2\iff a=2$$

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L'Hopital's rule can be used because we have a $\frac{0}{0}$ form, but the derivative of $\tan(ax)$ is not $\frac{1}{a \cos x}$.

If it helps, let $u = ax$, so $\frac{d}{dx} \tan(u) = \frac{d}{du} \frac{du}{dx} \tan(u) = \frac{d}{du} \tan(u) \cdot \frac{du}{dx}$ by the chain rule. Then the derivative is $\sec^2(u) \cdot a = a \sec^2(ax)$, using $u = ax$ again.

This gives (we can now substitute!) :

$$\lim_{x \to 0} \frac{a \sec^2(ax)}{\cos(x)} = \frac{a \sec^2(0)}{\cos(0)} = \frac{a \cdot 1}{1} = 2 \iff a=2.$$

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