Evaluate the following limits if it exist

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Homework Statement



evaluate the following limits if it exist
lim x sin1/x and limit x sin 1/x
x→0 x→∞

Homework Equations





The Attempt at a Solution


Someone have told me that I should let t=1/x and rewrite the limits.However, once I rewrite the limit, I still cannot evaluate the limits.
So,how can I find the limits?
 
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First rewrite the expression as $$\lim_{x \rightarrow 0} \frac{\sin\left(1/x\right)}{1/x}$$ and then the reason for introducing the variable ##t## becomes more obvious.
 
You should have learned the limit [tex]\lim_{\theta\to 0}\frac{sin(\theta)}{\theta}[/tex] in first semester Calculus. If you do not remember it, look in a Calculus text in the section on the derivative of sin(x).
 
haha1234 said:

Homework Statement



evaluate the following limits if it exist
lim x sin1/x and limit x sin 1/x
x→0 x→∞

Homework Equations





The Attempt at a Solution


Someone have told me that I should let t=1/x and rewrite the limits.However, once I rewrite the limit, I still cannot evaluate the limits.
So,how can I find the limits?
The OP's problem statement wasn't very clear, but there are two problems here.

$$\lim_{x \to 0} x sin(1/x)$$
$$\lim_{x \to \infty} x sin(1/x)$$

The hint applies to the second problem.
 
HallsofIvy said:
You should have learned the limit [tex]\lim_{\theta\to 0}\frac{sin(\theta)}{\theta}[/tex] in first semester Calculus. If you do not remember it, look in a Calculus text in the section on the derivative of sin(x).

I have used this method,but I still cannot find correct answer that I got the answer is 1.
 
CAF123 said:
First rewrite the expression as $$\lim_{x \rightarrow 0} \frac{\sin\left(1/x\right)}{1/x}$$ and then the reason for introducing the variable ##t## becomes more obvious.

Why limx→0xsin(1/x)is the same as limx→0sin(1/x)1/x ?:confused::shy:
 
haha1234 said:
Why limx→0xsin(1/x)is the same as limx→0sin(1/x)1/x ?:confused::shy:
It isn't the same. What you wrote is sin(1/x) * 1/x. What it should be is sin(1/x)/(1/x).