compute $\lim_{x\to 0}\frac{\sin(\tan(x))-\sin(\sin(x))}{\tan(\tan(x))-\tan(\sin(x))}$ 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.
 A: 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.
A: We can manipulate the given expression as follows
$$\frac{\sin(\tan(x))-\sin(\sin(x))}{\tan(\tan(x))-\tan(\sin(x))}=\frac{\frac1{\cos^3x }\frac{\sin(\tan(x))-\tan x}{\tan^3 x}-\frac{\sin(\sin(x))-\sin x}{\sin^3 x}+\frac{\tan x-\sin x}{\sin^3 x}}{\frac1{\cos^3x }\frac{\tan(\tan(x))-\tan x}{\tan^3 x}-\frac{\tan(\sin(x))-\sin x}{\sin^3 x}+\frac{\tan x-\sin x}{\sin^3 x}}$$
then we can use that as $x \to 0$
$$\frac{\sin x-x}{x^3}\to -\frac16$$
$$\frac{\tan x-x}{x^3}\to \frac13$$
with the methods indicated here:

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*Are all limits solvable without L'Hôpital Rule or Series Expansion
A: This answer will use some trigonometric identities and the standard limits $$\lim_{y\to 0}\frac{\sin y}y =1 \text{ and } \lim_{y\to 0}\frac y{\tan y} =1\tag 1$$
Using $\sin C-\sin D=2\cos \frac{D+C}2\sin \frac{C-D}2$, numerator of the given expression becomes: $$2\sin \Big(\frac{\tan x-\sin x}{2}\Big)\cos \frac{\tan x+\sin x}{2}\tag 2$$
Using $\tan A- \tan B=(1+\tan A\tan B)\tan (A-B)$, denominator of the given expression becomes $$(1+\tan(\tan x)\tan(\sin x))\tan (\tan x-\sin x)\tag 3$$
So re-writing the given expression using $(1), (2)$ and $(3)$ gives:
$$2\frac{\cos (\frac{\tan x+\sin x}{2})}{1+\tan(\tan x)\tan(\sin x)}.\frac{\sin \Big(\frac{\tan x-\sin x}{2}\Big)}{\frac{\tan x-\sin x}{2}}.\frac{\frac{\tan x-\sin x}{2}}{\tan x-\sin x}.\frac{\tan x-\sin x}{\tan (\tan x-\sin x)}\tag 4$$
By $(1)$, the first two terms in $(4)$ have limits (as $x\to 0$) $2$ and $1$ respectively and the last term has limit $1$ as $x\to 0$. The third term has limit equal to $\frac 12$ as $x\to 0$ so by limit rules the limit of the given expression is $2\times 1 \times \frac 12 \times 1=1$ as $x\to 0$.
A: Link to the answer image please check
The answer that i got for this problem is 1 without using L-Hospital or Maclurian series.
