How can I solve improper limits in calculus?

[frasifrasi]

Improper -- How do i do this?

I can't get started with the following improper limits:

lim as x approaches infinity of (x * sin (1/x))

I get this is in the form 0 * infinity, but when I move x to the bottom, and do l'hopital, nothing gets simpler.

And if you have a chance,

what is the lim as x approaches infinity of (cos x)/x ? I don't know what the limit of cos (infinity) is , so I can't get started. Thanks.

 

[BSMSMSTMSPHD]

cos(x) has no limit at infinity, because the function oscillates. But, it's bounded, and the denominator (x) is growing without bound. What will happen to the fraction as x gets larger?

As for the first one, try replacing 1/x with a new variable, say w. Then, the limit as z approaches infinity of x sin (1/x) is the same as the limit as w approaches 0 of sin(w)/w, a limit you should know!

 

[frasifrasi]

-->will it approach 0 since we take cos to be 1 or a treat it as any number?

--> for the first one, the answer is one--suing hospital or the formula sin t/t = 1.

 

[Dick]

Yes for the sin question. For the cos, -1/x<=cos(x)/x<=1/x. Squeeze.

 

[HallsofIvy]

You know, do you not, that \lim_{x\rightarrow 0} \frac{sin(x)}{x}= 1?

If so, a simple substitution should put x sin(1/x) in that form.