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.

To begin with, you split the limit into two ones that are $+\infty$, so the very first line is $\infty-\infty$. Even though those two infinities are comparable at the beginning (the limit exists) you cannot handle them separately as numbers and replace $\infty$ with $1\cdot\infty$. Note that all the operations with limits (sum, product, fraction) are justified only when all the limits exists and are finite. Take a simple example
$$
\lim_{x\to 0}\frac{\sin x-x}{x^3}=-\frac{1}{6}\quad (\text{L'Hospital two times}),
$$
but if you do what you did
$$
\lim_{x\to 0}\frac{\sin x-x}{x^3}=\lim_{x\to 0}\frac{\sin x}{x^3}-\lim_{x\to 0}\frac{x}{x^3}=\underbrace{\lim_{x\to 0}\frac{\sin x}{x}}_{=1}\lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2}
$$
you will get nonsense.
In your example, you should keep both fractions under one limit and split them only when you are sure that the limits are finite. You may continue as
$$
\lim\frac{\tan^2x-x^2}{5\sin^4x\cos x}=\lim\frac{\tan x-x}{\sin^3 x}\cdot\lim\frac{\tan x+x}{5\sin x\cos x}
$$
or rewrite at the beginning as
$$
\lim_{x \rightarrow 0}\frac{\tan x- x-\frac{x^3}{3}}{\sin^5x}=
\lim_{x \rightarrow 0}\frac{\sin x- x\cos x-\frac{x^3}{3}\cos x}{x^5}\cdot
\lim_{x\to 0}\frac{x^5}{\sin^5x\cos x}.
$$
The last limit in the product in both cases is easily calculated.
