# Solve limit for a

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?

• Are you sure the limit is to $\infty$, not to $0$? – Toby Mak Apr 11 at 7:21
• @TobyMak you're right, my bad! – Julius Apr 11 at 7:53

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: $$\lim_{x\to0}\frac{\tan(ax)}{\sin(x)}=\frac{\tan0}{\sin0}=\frac{0}{0}\stackrel{\text{l'H}}{=}\lim_{x\to0}\frac{\frac{a}{\cos^2(ax)}}{\cos(x)}=\lim_{x\to0}\frac{a}{\cos^2(ax)\cos(x)}=\frac{a}{\cos^20\cos0}=a.$$
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 $$\lim_{x\to 0}\frac{\tan{ax}}{\sin{x}}=\lim_{x\to 0} \frac{ax+\frac{(ax)^3}{3}+o(x^3)}{x-\frac{x^3}{6}+o(x^3)}=\lim_{x\to 0}\frac{ax}{x}=a=2\iff a=2$$
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.
$$\lim_{x \to 0} \frac{a \sec^2(ax)}{\cos(x)} = \frac{a \sec^2(0)}{\cos(0)} = \frac{a \cdot 1}{1} = 2 \iff a=2.$$