Find the limit of $\lim\limits_{x\rightarrow0}\frac{x}{\tan x}$. Find the limit of $\lim\limits_{x\rightarrow0}\frac{x}{\tan x}$

Clearly, since the limit takes the form of $\frac{0}{0}$, one should try L'Hopital's Rule. If we apply L'Hopital's Rule, the problem is that $\frac{d(\tan x)}{dx}=\cot x$, and $\cot(0)$ is undefined. Thus, we cannot find a limit using L'Hopital's Rule.
My problem set suggests that this question can be answered using little more than L'Hopital and Cauchy's Mean Value Theorem. Not sure how to proceed. Please help.
 A: No, the derivative of $\tan x$ with respect to $x$ is $\sec^2x$, not $\cot x$. But you don’t need l’Hospital’s rule:
$$\frac{x}{\tan x}=\frac{x}{\sin x}\cdot\cos x\;,$$
and you should know the limits of both factors as $x\to 0$.
A: Hint: $\lim_{x \to 0} \dfrac{x}{\sin x} = 1$ and $\dfrac{x}{\tan x} = \dfrac{x} {\sin x} \ \cos x$.
A: Notice 
$$ \frac{x}{ \tan x}  = \frac{ \cos x}{\sin x} x \to 1 \times 1  = 1 $$ as $x \to 0 $ since $\frac{x }{\sin x} \to 1 $ as $x \to 0 $
A: $\lim_{x\rightarrow 0} \frac{x} {\tan x}=\lim_{x\rightarrow 0} \frac{x\cos x} {\sin x}=\lim_{x\rightarrow 0} \frac{\cos x} {\frac{\sin x}{x}}=1$
A: Here are the steps
$$ \lim\limits_{x\to 0} \frac{x}{\tan x} = \lim\limits_{x\to 0} \frac{x}{\frac{\sin x}{\cos x}} = \lim\limits_{x\to 0} \frac{x\cos x}{\sin x}$$
$$ = \left(\lim\limits_{x\to 0} \frac{x}{\sin x}\right)\left(\lim\limits_{x\to 0} \cos x\right) = 1\cdot 1 = 1$$
Note that we can split the original limit into two convergent limits, as shown above.
A: Hope you know that $(\sin x)'=\cos x,(\cos x)'=\sin x$ and differentiating a fraction of functions. This will be very useful in your academic life:
$$\frac{d}{dx}\tan x=\frac{d}{dx}\frac{\sin x}{\cos x}=\frac{\cos x.\cos x-\sin x.(-\sin x)}{\cos^2x}=\frac{\cos^2x+\sin^2x}{\cos^2x}=\sec^2x$$
So:
$$\lim_{x\to0}\frac{x}{\tan x}=\lim_{x\to0}\frac{1}{\sec^2x}=\frac{1}{1^2}=1$$
