Compute the limit without L'Hospital's rule I'm not sure how to handle the trig functions with different arguments when computing this limit using L'Hospital's rule. 
$$\lim_{x \rightarrow 0} \frac {x^2\cos(\frac {1} {x})} {\sin(x)}.$$
I have come up with the correct numerical answer via a different method, but am unsure if the logic would hold true for all cases (maybe I arrived at the correct answer by chance).
Here is my working:

Let $g(x)=x^2\cos(\frac 1 x)$ and $h(x) = \frac 1 {\sin x} = \csc x$.
We know that the following holds true for all $x$: $$-1 \le \cos (\frac 1 x) \le 1$$
  Since $x^2 \ge 0$ for all x:
  $$-x^2 \le x^2\cos (\frac 1 x) \le x^2$$
  Taking limits as $x \rightarrow \infty$ gives:
  $$ \lim_{x\rightarrow 0}(-x^2)\le \lim_{x\rightarrow 0}(x^2\cos (\frac 1 x)) \le \lim_{x\rightarrow 0}(x^2)$$
  By the sandwich rule (or squeeze theorem):
  $$ 0 \le  \lim_{x\rightarrow 0}(x^2\cos (\frac 1 x)) \le 0$$
  $$ \Rightarrow \lim_{x\rightarrow 0}(x^2\cos (\frac 1 x)) $$
  And hence, due to the algebra of limits:
  $$\lim_{x \rightarrow 0} \frac {x^2\cos(\frac {1} {x})} {\sin(x)} = 0.$$

 A: $$
\frac {x^2\cos\frac 1 x} {\sin x} =  x\cdot \frac x {\sin x} \cdot \cos\frac 1 x
$$
Now use the fact that $-1 \le \cos \frac 1 x \le 1$ and $x\to0$ and one further fact not mentioned in your question:
$$
\frac x {\sin x} \to 1 \text{ as } x\to0.
$$
Without that last fact or something else other than what's in your question, you haven't dealt with the fact that $\sin x \to0.$
A: Note that $$\lim_{x \to 0}\frac{\sin(x)}{x} = \lim_{x \to 0}\frac{x}{\sin(x)} = 1$$
Then
\begin{align}
L=\lim_{x \to 0} \frac{x^2\cos(1/x)}{\sin(x)} &= \lim_{x \to 0}\frac{x}{\sin(x)}\frac{x \cos(1/x)}{1}\\
&=\lim_{x \to 0}x \cos(1/x)
\end{align}
And by Squeeze Theorem
$$-1 \le \cos(1/x) \le1$$
$$-x \le x\cos(1/x) \le x$$
$$0 \le \lim_{x \to 0}x \cos(1/x) \le 0$$
$$L=0$$
A: $$\begin{align}L&=\lim_\limits{x\to0}\dfrac x{\sin x}\cdot\lim_\limits{x\to0}x\cos\left(\dfrac1x\right)\\&=\lim_\limits{x\to0}x\cos\left(\dfrac1x\right)\end{align}$$
We know
$$\begin{align}-1&\le\cos\left(\dfrac1x\right)\le1\\-x&\le x\cos\left(\dfrac1x\right)\le x\\\lim_\limits{x\to0}(-x)&\le\lim_\limits{x\to0} x\cos\left(\dfrac1x\right)\le\lim_\limits{x\to0}x\end{align}$$
So, $$\lim_\limits{x\to0} x\cos\left(\dfrac1x\right)=0$$ via the Squeeze Theorem
And hence $$L=0$$ via $$\lim_{n\to c} a_nb_n=\lim_\limits{n\to c}a_n\cdot\lim_\limits{n\to c}b_n$$
A: Use asymptotic analysis: you know $\sin x \sim_0 x$, hence $\dfrac{x^2}{\sin x}  \sim_0 \dfrac{x^2}x=x $. 
Furthermore, $\Bigl\vert\cos\dfrac1x \Bigr\vert \le 1$, so
$$\smash{\dfrac{x^2\cos\dfrac1x}{\sin x}} = O(x)\to 0.$$
