Proving the Limit of x*sin(x) as x Approaches Infinity is Equal to 1

[karkas]

Homework Statement

The complete exercise is:

If $\lim_{x->\inf } \frac{f(x)-5x^2sin(x)}{(\sqrt (x^2+2))-x} = 7$

show that $\lim_{x->\inf} \frac{f(x)}{x} = 5$

Homework Equations

How do I show that $\lim_{x->\inf} xsinx =1$, because I run into it!

The Attempt at a Solution

I set K(x) = the fraction of the first limit and I solved for f(x) (x=0 excluded).

Then I have the limit $\lim_{x->\inf} \frac{f(x)}{x} = \lim_{x->\inf} K(x)*0 + 5 xsinx$.

Yet finally I reach the limit I spoke about in 2.

 

[zcd]

You can't, it eventually oscillates between +/- infinity. What exactly is f(x) in this context?

 

[karkas]

Random function. It doesn't specify... Any other solutions?

 

[zcd]

Maybe you're omitting part of the question?
If $\lim_{x->\inf } \frac{f(x)-5x^2sin(x)}{(\sqrt (x^2+2))-x}$
doesn't say anything because you're only giving the condition. Does the limit = something? Is the question asking you to find f(x) such that $\lim_{x->\inf} \frac{f(x)}{x} = 5$?

 

[karkas]

Yes indeed I'll fix it.

No, it just wants me to prove the second limit equals 5.