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?
 A: 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
$$\frac{\sin 5x-5\sin 3x+10\sin x}{\sin (\sin x)+\tan x-2x}\sim_0\frac{16s^5}{\dfrac {7s^5}{30}}=\frac{480}7.$$
A: The solution via Taylor series is indicated in my comments and perfectly handled in another answer. In comments to question OP asks for a solution without the use of Taylor series and I don't think it is possible to avoid both Taylor and L'Hospital's Rule (indicated by tag of the question). Here is an approach via L'Hospital's Rule (also suggested in my comments).
We will evaluate the limit $$L=\lim_{x\to 0}\frac{\sin(\sin x) +\tan x - 2x}{x^5}$$ and the answer to the current question is $16/L$. Let's first use the substitution $\sin x=t, \tan x=t(1-t^{2})^{-1/2}$ so that $t\to 0$ and $t/x\to 1$ and thus the above limit is transformed into $$\lim_{t\to 0}\frac{\sin t +t(1-t^{2})^{-1/2}-2\arcsin t} {t^{5}}$$ and this we evaluate below via L'Hospital's Rule.
\begin{align}
L&=\lim_{t\to 0}\frac{\cos t+(1-t^{2})^{-3/2}-2(1-t^{2})^{-1/2}}{5t^{4}}\text{ (via L'Hospital's Rule)} \notag\\
&=\frac{1}{5}\lim_{t\to 0}\frac{(1-t^{2})^{3/2}\cos t-(1-2t^{2})}{t^{4}(1-t^{2})^{3/2}}\notag\\
&=\frac{1}{5}\lim_{t \to 0}\frac{(1-t^{2})^{3}\cos^{2}t-(1-2t^2)^{2}}{t^{4}\{(1-t^{2})^{3/2}\cos t+(1-2t^2)\}}\notag\\
&=\frac{1}{10}\lim_{t\to 0}\frac{(1-3t^2)\cos^{2}t-1+4t^2}{t^4}+\frac{3t^4\cos^{2}t-t^{6}\cos^{2}t-4t^{4}}{t^4}\notag\\
&=\frac{1}{10}\left(-1+\lim_{t\to 0}\frac{t^2-\sin^{2}t+3t^{2}\sin^{2}t}{t^4}\right)\notag\\
&=\frac{1}{10}\left(2+\lim_{t\to 0}\frac{t+\sin t} {t} \cdot\frac{t-\sin t} {t^3}\right)\notag\\
&=\frac{1}{10}\left(2+2\lim_{t\to 0}\frac{t-\sin t} {t^3}\right)\notag\\
&=\frac{1}{5}\left(1+\lim_{t\to 0}\frac{1-\cos t} {3t^2}\right)\text{ (via L'Hospital's Rule)} \notag\\
&=\frac{1}{5}\cdot\left(1+\frac{1}{3}\cdot\frac{1}{2}\right)\notag\\
&=\frac{7}{30}\notag
\end{align}
The desired limit in question is thus $16/L=480/7$.

The substitution $t=\sin x$ makes our life easy for both L'Hospital's Rule and Taylor series (see the other answer here by Bernard). Often a direct application of these tools is unwieldy and it is preferable to apply some substitution or algebraic manipulation before resorting to these tools. 
