What's the right idea to evaluate $\lim\limits_{x\to 0} \sqrt{x}\sin(x^n)/ (\sin(x))^m$?

Question

The problem: $\lim\limits_{x\to 0}\frac{\sqrt{x}\sin(x^n)}{(\sin(x))^m}$

First step - it's $0/0$, so I decided to use L'Hospital's rule, but encountered with $0/0$ again and again.

Then i tried to use $\lim\limits_{x\to0}\sin(x)/x = 1$ by creating $x^m$ and $x^n$ but also stuck on $0/0$

After that remained Taylor series, but I suppose there is a much easier solution.

Answer 1

Multiply and divide up and down by $x^m$ and by $x^n$ $$ \lim_{x\to0} \sqrt{x}x^{n-m}\frac{\sin(x^n)}{x^n}\left(\frac{x}{\sin x}\right)^m $$ and using $\sin(x)/x\to 1$ as $x\to 0$, the limit is effectively equivalent to the limit as $x\to 0$ of $x^{n-m+1/2}$, which is $0$ if $n-m+1/2>0$, $1$ if $n-m+1/2=0$ and $\infty$ if $n-m+1/2<0$.

Answer 2

$$\lim _{ x\to 0 } \frac { \sqrt { x } \sin (x^{ n }) }{ (\sin (x))^{ m } } =\lim _{ x\to 0 } \frac { \sin (x^{ n }) }{ x^{ n } } \cdot \frac { x^{ m } }{ (\sin (x))^{ m } } { x }^{ n-m+1/2 }=\lim _{ x\rightarrow 0 }{ { x }^{ n-m+\frac { 1 }{ 2 } } } =\begin{cases} \infty ,\quad n-m+1/2<0 \\ 1,n-m+1/2=0 \\ 0,n-m+1/2>0 \end{cases}$$

Answer 3

The solution involving Taylor series is not really complicated. Just remember that $\sin t = t + o(t^2)$ when $t \to 0$ and you can rewrite the limit as $$\lim_{x \to 0} \frac{\sqrt{x}x^n}{x^m} = \lim_{x \to 0} x^{n - m + 1/2} = \ldots$$

Now distinguish the different cases based on the value of $n - m + 1/2$ and you can evaluate the limit.