Question

Question #597c8

Answer 1

The limit is 1.

Explanation:

`lim_(xrarroo)x/(x+sinx) = lim_(xrarroo)1/(1+(sinx/x))`

Note that `lim_(xrarroo)sinx/x = 0` (by the squeesze theorem).

So

`lim_(xrarroo)x/(x+sinx) = lim_(xrarroo)1/(1+(sinx/x))`

`= 1/(1+0) = 1`

Note on l'Hospital's Rule

Although the initial form is `oo/oo`, when we try to apply l'Hospita'l, we get

`lim_(xrarroo) 1/(1+cosx)`.

This limit does not exist, but it is not infinite. Therefore, l'Hospital gives us no information about the limit.