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Answer 1 · 2017-01-08T16:46:55

The limit is 1.

#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.