Finding $\lim_{x\rightarrow 0}\frac{\sin^n(x)}{x}$ Finding $$\lim_{x\rightarrow 0}\frac{\sin^n(x)}{x}$$ when $n$ is a positive integer number
I think this limit equal to 1.How can I find this limit? 
 A: $$\lim_{x \to 0} \frac{\sin^n x}{x} = \lim_{x \to 0} \frac{\sin^n x}{x^n} x^{n-1} = \left(\lim_{x \to 0} \frac{\sin x}{x} \right)^n  \cdot \lim_{x \to 0} x^{n-1} = 0$$
A: rewrite it in the form $\lim_{ x\to 0}\frac{\sin(x)}{x}\sin^{n-1}(x)$
A: For $n>1$, the limit is zero, because $\sin x \sim x$ at $x\to 0$, so you have something that behaves like $x^{n-1}$.
A: It depends on what $n$ is.
If $n>1$, the answer is different from when $n = 1$. Do you know the rule of l'Hopital?
A: Since $\sin x \sim x$ as $x\to 0$, then the limit is equal to $\lim_{x\to 0}x^{n-1}$.
So, if $n < 1$, then the limit does not exist; if $n=1$, then the limit is $1$; and $n>1$, then clearly the limit is $0$.
A: First note that
$$ \lim_{x\to 0}\frac{\sin x}{x}=1 $$
And 
$$ \lim_{x\to 0} \sin x= \lim_{x\to 0} x $$
So now we have 
$$ \lim_{x\to 0}\frac{\sin^n x}{x}= \lim_{x\to 0} \frac{x^n}{x} = \lim_{x\to 0} x^{n-1} $$
Therefore 
$$ \lim_{x\to 0} x^{n-1}=0, \quad \forall n\in\mathbb{R}:n\gt 1$$
$$ \lim_{x\to 0} x^{n-1}=1, \quad \forall n\in\mathbb{R}:n= 1$$
$$ \lim_{x\to 0} x^{n-1}=\infty, \quad \forall n\in\mathbb{R}:n\lt 1,n\ \mbox{is odd} $$
$$ \lim_{x\to 0} x^{n-1}=\mbox{does not exist}, \quad \forall n\in\mathbb{R}:n\lt 1,n\ \mbox{is even}$$
