<s>[INST] Evaluate $\lim_{x \rightarrow 0}\frac{\tan x- x-\frac{x^3}{3}}{\sin^5x}$ How do I ind the limit of 
$$\lim_{x \rightarrow 0}\frac{\tan x- x-\frac{x^3}{3}}{\sin^5x}$$
by L'Hopital's Rule?
Using Desmos, I get the answer that this limit evaluates to $\frac{3}{15}$, but I can't get that answer.
This is what I've done so far:
\begin{align}
&\lim_{x \rightarrow 0}\frac{\tan x -x}{\sin^5x}-\lim_{x \rightarrow 0}\frac{x^3/3}{\sin^5x}\\
&=\lim_{x \rightarrow 0}\frac{\sec^2x-1}{5\sin^4x\cos x}-\lim_{x \rightarrow 0}\frac{x^2}{5\sin^4x\cos x}\\
&=\lim_{x \rightarrow 0}\frac{\tan^2x}{5\sin^4x\cos x}-\lim_{x \rightarrow 0}\frac{x^2}{\sin^2x}\lim_{x \rightarrow 0}\frac{1}{5\sin^2x\cos x}\\
&=\lim_{x \rightarrow 0}\frac{1}{5\sin^2x\cos^3x}-\lim_{x \rightarrow 0}\frac{\cos^2x}{5\sin^2x\cos^3 x}\\
&=\frac{1}{5}
\end{align}
I checked by answer with Desmos, and the 4th line is where the difference occurs. In the 4th line, I did this.
$$
\lim_{x \rightarrow 0}\frac{x^2}{\sin^2x}=\lim_{x \rightarrow 0}\frac{1}{(\sin x/x)^2}=1
$$
Shouldn't this be a valid operation? I'm just using the product law on limits. If anyone could tell me where I've gone wrong, I will be grateful.
 [/INST] As noted by А.Г., you can not split the limit unless both exist. 
You can use L'Hospital right away and see how splittings are done:
$$\lim_{x \rightarrow 0}\frac{\tan x- x-\frac{x^3}{3}}{\sin^5x}\stackrel{LR}=
\lim_{x \rightarrow 0}\frac{\sec ^2x- 1-x^2}{5\sin^4x\cdot \cos x}=\\
\lim_{x \rightarrow 0}\frac{1-\cos^2x-x^2\cos^2x}{5\sin^4x}\cdot \underbrace{\lim_{x\to 0}\frac1{\cos^3x}}_{=1}\stackrel{LR}=\\
\lim_{x \rightarrow 0}\frac{\sin 2x-2x\cos^2x+x^2\sin 2x}{20\sin^3x}\cdot \underbrace{\lim_{x\to 0}\frac{1}{\cos x}}_{=1}\stackrel{LR}=\\
\lim_{x \rightarrow 0}\frac{2\cos 2x-2\cos^2x+4x\sin 2x+2x^2\cos 2x}{60\sin^2x}\cdot \underbrace{\lim_{x\to 0}\frac{1}{\cos x}}_{=1}=\\
\lim_{x \rightarrow 0}\frac{-2\sin^2x}{60\sin^2x}+\lim_{x \rightarrow 0}\frac{4x\sin 2x}{60\sin^2x}+\lim_{x \rightarrow 0}\frac{2x^2\cos 2x}{60\sin^2x}=\\
-\frac1{30}+\frac2{15}+\frac1{30}=\frac2{15}.$$
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