Question

Calculate:
\[\lim_{x \rightarrow +\infty} 2 \frac{\sin x}{x} + \frac{x}{2} \sin \frac{1}{x}\]

Answer 1

The first term goes to zero, right? In the second term \[\frac{x}{2}\] goes to infinity and \[\sin \frac{1}{x}\] goes to zero. So I get infinity times zero. Then...?

Answer 2

Look up L'Hopital's rule.

Answer 3

Ok, I differentiated both numerator and denominator and got the following:
\[\lim_{x \rightarrow +\infty} \frac{\sin{\frac{1}{x}}}{\frac{2}{x}} = \lim_{x \rightarrow +\infty} \frac{-\frac{1}{x^2} \times \cos{\frac{1}{x}}}{-\frac{2}{x^2}} = \lim_{x \rightarrow +\infty} \frac{\cos{\frac{1}{x}}}{2} = \frac{1}{2}\]
Is it right?

Answer 4

Yep, looks like it :)

Answer 5

Hooray!

Answer 6

Well done, mate ;)