Question

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

Answer 1

L'Hopital's Rule only applies if the limit is the indeterminate `0/0`

Explanation:

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

Answer 2

There is one error and an additional explanation is needed for why it won't work even if we correct the error. I have added a second edit .

Explanation:

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`