Question #597c8

1 Answer
Jan 8, 2017

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.