Compute $\lim_{x\to 0}\frac{\sin(\tan(x))-\sin(\sin(x))}{\tan(\tan(x))-\tan(\sin(x))}$

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Compute $\lim_{\theta\rightarrow 0}\frac{\sin{(\tan{\theta})}-\sin{(\sin{\theta})}}{\tan{(\tan{\theta})}-\tan{(\sin{\theta})}}$

(4 answers)

Closed 2 years ago.

I need help with this problem as stated in the title:

$$\lim_{x\to 0}\frac{\sin(\tan(x))-\sin(\sin(x))}{\tan(\tan(x))-\tan(\sin(x))}$$

I'm trying to convert the problem to using only the standard limits, with that said, I'm trying to solve the problem without using L'Hospital nor any Maclaurin series.

Do you guys have any tips on how to tackle the problem as I'm really struggling with this one.

Thank you.

Answer 1

Alternate approach:

\begin{align} \frac{\sin(\tan x)-\sin(\sin x)}{\tan(\tan x)-\tan(\sin x)} &= \frac{2\sin(\frac{\tan x-\sin x}{2})\cos(\frac{\tan x+\sin x}{2})}{\frac{\sin(\tan x-\sin x)}{\cos(\tan x)\cos(\sin x)}} \\ &= \frac{2\sin(\frac{\tan x-\sin x}{2})\cos(\frac{\tan x+\sin x}{2})}{2\sin(\frac{\tan x-\sin x}{2})\cos(\frac{\tan x-\sin x}{2})}\cdot \cos(\tan x)\cos(\sin x) \\ &= \frac{\cos(\frac{\tan x+\sin x}{2})}{\cos(\frac{\tan x-\sin x}{2})}\cdot \cos(\tan x)\cos(\sin x) \end{align}

and its limit as $x\to0$ can be evaluated easily.