$\lim\limits_{x\to\infty}x^a\sin(1/x)$ $$\lim_{x \rightarrow \infty} x^a\sin{\frac{1}{x}}$$
for this limit ,it was showed on the textbook that
$$\lim_{x \rightarrow \infty} x^a\sin{\frac{1}{x}}=\begin{cases} 0 & a<1 \\ 1 & a=1 \\ \infty & a>1 \end{cases}$$
in my opinion ,it's obviously that $\lim\limits_{x \rightarrow \infty}\sin{\frac{1}{x}}=0$ and that $\lim\limits_{x \rightarrow \infty}x^a=\infty$ 
I wonder what's wrong with my view
 A: There is no need for L'Hospital here if you already know that $\lim_{x\to0}\frac{\sin x}{x}=1$.
From the above identity we can get the following:
$$\lim_{x\to\infty}x^a\sin\frac1x=\lim_{x\to\infty}x^{a-1+1}\sin\frac1x=\lim_{x\to\infty}x^{a-1}x\sin\frac1x=\lim_{x\to\infty}x^{a-1}\frac{\sin\frac1x}{\frac1x}$$
If $a=1$ then we have the limit above is equal to 1 ($x^0=1$ and set $t=\frac1x$, now you have $\frac{\sin t}t$ as $t\to 0$).
If $a<1$ then we have $x^{a-1}=\frac1{x^{1-a}}$ approaches zero, and we have a multiplication of a bounded limit by a zero limit. Thus zero.
I $a>1$ then $a-1>0$ and so $x^{a-1}$ approaches $\infty$ and the $\sin$ part is finite non-zero. So the limit is infinity.
A: The intuition is that in certain cases the first one may go to zero more quickly than the second one goes to infinity or visa-versa. 
The way to answer this question is to use l'hospital rule : if $a < 0$, then the answer is clear. If $a > 0$, apply l'hospital rule to
$x^a\sin \frac{1}{x} = \frac{\sin \frac{1}{x}}{x^{-a}}$. 
Then consider what happens for $0 < a < 1$, $a = 1$ and $a > 1$. 

Doing the calculation, you get that the limit is
$\lim_{x \rightarrow \infty} \frac{-x^{-2}\cos\frac{1}{x}}{-ax^{-a - 1}} = x^{a - 1}\cos\frac{1}{x} = \cos(0)\lim_{x \rightarrow \infty} x^{a - 1}$
$=\begin{cases}
0 & \quad 0 < a < 1 \\
1 & \quad a = 1 \\
\infty & \quad a > 1
\end{cases}$
A: $\sin\frac{1}{x}\sim_\infty\frac{1}{x}$, so $x^a\sin\frac{1}{x}\sim_\infty x^{a-1}$, so you have your result.
A: put $y=1/x$, then your problem becomes $\lim_{y \to 0^+} \frac{\sin(y)}{y^a}$.
Then expand $\sin(y)$ in Taylor series form, divide that series by $y^a$ and check the limit for domain of $a$ or you can just apply L'Hopital Rule. you will get the final result.
