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$$\lim_{x \rightarrow \infty} x^a\sin{\frac{1}{x}}$$

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

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@Francis King: math.stackexchange.com/faq#etiquette –  Aang Jun 23 '12 at 10:05
    
oh thanks for reminding.i have just get the way to click accept. –  Francis King Jun 23 '12 at 10:14
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4 Answers

up vote 4 down vote accepted

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.

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thanks. this one is just i want. –  Francis King Jun 24 '12 at 16:03
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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

$\lim_{x \rightarrow \infty} \frac{-x^{-2}\cos\frac{1}{x}}{-ax^{-a - 1}} = x^{a - 1}\cos\frac{1}{x} = \cos(0)\lim_{x \rightarrow \infty} x^{a - 1}$

$=\begin{cases} 0 & \quad 0 < a < 1 \\ 1 & \quad a = 1 \\ \infty & \quad a > 1 \end{cases}$

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thanks. your way to use l'hospital rule makes me clear.but this question is founded in section before l'hospital rule was taught.i wonder whether there should be other ways with essential method such as definition of limit –  Francis King Jun 23 '12 at 10:04
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$\sin\frac{1}{x}\sim_\infty\frac{1}{x}$, so $x^a\sin\frac{1}{x}\sim_\infty x^{a-1}$, so you have your result.

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First point is not correct. If $a = \frac{1}{2}$, $\lim_{x \rightarrow \infty} \sqrt{x} \neq 0$. –  William Jun 23 '12 at 9:31
    
I just woke up and I'm under caffeinated :-p. Corrected. –  JBC Jun 23 '12 at 9:38
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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.

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