How do you use l'Hopital's Rule to evalute #lim_(xrarroo) (x+sin(x))/x#?

#lim_(xrarr00)(x * sin(x))/x != lim_(xrarroo)0/0#

Error

#d/dx(xsinx) = sinx+xcosx# #" "# (Use the product rule.)

Corrected version

#lim_(xrarroo)(xsinx)/x# has initial form #oo/oo#.

We apply l"Hopital's Rule for #f(x)/g(x)# and attempt to find

#lim_(xrarroo)(sinx+xcosx)/1#.

This limit does not exist and is neither #oo# nor #-oo#, therefore l'Hopital does not apply.

Explanation

In order to use

#lim_(xrarra)f(x)/g(x) = lim_(xrarra)f'(x)/g'(x)#

we must have one of the appropriate initial indeterminate forms (#0/0# or #(+-oo)/(+-oo)#)

and also we must have

# lim_(xrarra)(f'(x))/(g'(x))# is finite or #oo# or #-oo#.

In this question, # lim_(xrarra)(f'(x))/(g'(x))# is not helpful.

Second Edit

It occurs to me that there may be a typographic error in the question.

It may be that the intended question involved #f(x)/g(x) = (x+sinx)/x#.

Is so, then the derivative stated in the question is correct, but the explanation of why l'Hopital fails is the same.

In order to get an anwer from l'Hopital we must have

# lim_(xrarra)(f'(x))/(g'(x))# is finite or #oo# or #-oo#..

But #lim_(xrarroo)(1+cosx)# does not satisfy this condition.

Therefore l'Hopital's rule offers us no help for evaluating this limit.

(Note: This example shows the l'Hopital will not give answers for every indeterminate limit. Sometimes it fails.)

Finally, note that we can evaluate the limit without l'Hopital using #(x+sinx)/x = 1+sinx/x# which goes to #1+0 = 1# as #xrarroo#