<s>[INST] Find the limits of find the limits :
$$\lim_{x\to 0}\frac{\sin 5x-5\sin 3x+10\sin x}{\sin (\sin x)+\tan x-2x}$$
i know that :
$$\sin 5x=5\sin x-20\sin^3x+16\sin^5x$$
$$5\sin 3x=15\sin x-20\sin^3x$$
so we have :
$$\sin 5x-5\sin 3x+10\sin x=16\sin^5x$$
$$\lim_{x\to 0}\frac{16\sin^5x}{\sin (\sin x)+\tan x-2x}=\frac{16\sin^5 x}{x^5}\cdot\frac{x^5}{\sin(\sin x)+\tan x-2x}$$
$$\lim_{x\to0}\frac{(\sin (\sin x)-\sin x)+(\sin x-x)+(\tan x-x)}{x^5}=0$$
thus :
$$\lim_{x\to 0}\frac{\sin 5x-5\sin 3x+10\sin x}{\sin (\sin x)+\tan x-2x}=\infty$$
it is right?
 [/INST] Use  Taylor's formula at order $5$:
\begin{align}
&\frac{\sin 5x-5\sin 3x+10\sin x}{\sin (\sin x)+\tan x-2x}=\\[1ex]
&\frac{5x-\frac{125x^3}6+\frac{625x^5}{24}+o(x^5)-5\Bigl(3x-\frac{27x^3}6+\frac{81x^5}{40}+o(x^5)\Bigr)+10\Bigl(x-\frac{x^3}6+\frac{x^5}{120}+o(x^5)\Bigr)}{\sin\Bigl(x-\frac{x^3}6+\frac{x^5}{120}+o(x^5)\Bigr)+x+\frac{x^3}3+\frac{2x^5}{15}+o(x^5)-2x}\\[1ex]
=&\frac{16x^5+o(x^5)}{x-\frac{x^3}3+\frac{x^5}{10}+o(x^5)+x+\frac{x^3}3+\frac{2x^5}{15}+o(x^5)-2x}=\frac{16x^5+o(x^5)}{\frac{7x^5}{30}+o(x^5)}\\[1ex]
&=\frac{16+o(1)}{\frac7{30}+o(1)}\to\frac{16}{\frac7{30}}=\color{red}{\frac{480}{7}}.
\end{align}
Added (following @ParamanandSingh's strategy):
Set $s=\sin x$. As the O.P. pointed, Chebyshev' polynomials yield the formula
$$\sin 5x-5\sin 3x+10\sin x=16s^5.$$
On the other hand, in the denominator, we have $\tan^2x=\dfrac{s^2}{1-s^2}$, so in $\bigl(-\frac\pi2,\frac\pi2\bigr)$,
$$\tan x=\frac s{\sqrt{1-s^2}}=s\Bigl(1+\frac{s^2}2+\frac{3s^4}8+o(s^4)\Bigr).$$
Also, we have
$$x=\arcsin s=s+\frac12\, \frac{s^3}3 +\frac{1\cdot 3}{2\cdot 4}\,\frac{x^5}5$$
whence the following expansion at order $5$ for the denominator:
\begin{alignat}{2}
\sin(\sin x)+\tan x-2x&= \sin s+\frac s{\sqrt{1-s^2}}-2\arcsin s&
&=s -\frac{s^3}6+\frac{s^5}{120} \\[1ex]
&&&+s+\frac{s^3}2+\frac{3s^5}8 \\[1ex]
&&&-2s-\frac{s^3}3-\frac{3s^5}{20}+o(s^5)\\
&=\frac {7s^5}{30} +o(s^5),
\end{alignat}
and finally
$$\frac{\sin 5x-5\sin 3x+10\sin x}{\sin (\sin x)+\tan x-2x}\sim_0\frac{16s^5}{\dfrac {7s^5}{30}}=\frac{480}7.$$
</s>