Verified answer

No. L’Hopital’s rule, oo/oo version, says:

Suppose f(x) and g(x) goes to oo as x --> oo and that lim x--> oo f'(x)/g'(x) = L (L = oo and L = -oo are allowed). Then, lim x --> oo f(x) /g(x) = L.

Let f(x) = x + sin(x) and g(x) = 2x – 5sin(x). Then, it's true that f(x) and g(x) --> oo as x --> oo. But

f'(x) = 1 + cos(x) and g'(x)= 2 - 5 cos(x), so that ,

f'(x) /g'(x) = (1 + cos(x))/( 2 - 5 cos(x)). We readily see f'/g' doesn't not have a limit as x --> oo (it's periodic and non constant, for cos(x) <> 2/5). So, L’Hopital’s rule doesn't apply here.

Anyway, we readily see the limit is 1/2. I won't show why because I know you know why.

EDIT: Interesting to remark that L’Hopital’s rule could be used here if, instead of limit at oo, we had limit at 0. Then, L'Hopital would give the correct answer, -2/3.

EDIT(2)

No, the required condtions are NOT met in the oo/oo case. Like I wrote, a necessary condition is that lim x --> oo f'(x)/g'(x) = L, where L is a real number or oo, or -oo. In your case, this is NOT satisfied, so L'Hopital does not apply.

There's no bug, it's just that the required conditions are not satisfied in your case.

Corrected from earlier, apologies it was a very good question (have a star):

The L'Hopital rule is to be used when we want to find:

Lim f(x) / g(x) → m

x→a

but

where m is an indeterminant form such as 0/0, infinity/infinity or others on the reference.

But your example meets this critera:

Lim x-sin(x) → Lim x (as x >> 1) →∞

x→∞ x→∞

Lim 2x - 5sin(x) → Lim 2x (as 2x >> 5) →∞

x→∞ x→∞

i.e.

Lim [x + sin(x)] / [2x – 5sin(x)] →∞/∞

x→∞

which is an indeterminant form.

However a further condition of the rule however is that:

Lim f'(x) / g'(x) → n

x→a

exists

In your case we have:

Lim f'(x) / g'(x) = Lim [1 + cos(x)] / [2 – 5cos(x)]

x→∞ x→∞

Which has no limit and at points where cos(x) = 2/5 a zero denominator (and you can nor use L'Hopital as the limit of the ratio is not indeterminant)

The solution method to use is:

Lim [x + sin(x)] / [2x – 5sin(x)] =

x→∞

Lim [x / 2x] =

x→∞

Lim 1/2 = 1/2

x→∞

As demostrated by another answer if you used L’Hopital’s rule twice withot considering the conditions for application you just end up with:

Lim [1 + cos(x)] / [2 – 5cos(x)] =

x→∞

Lim [-sin(x)] / [– 5sin(x)] = 1/5

x→∞

which is wrong.

I do not agree with people who said yes, because x→∞

sinx does not have a limit nor does cos x.

EDIT: No doctor D its not a bug, because the if part of your condition is not met.

If f and g both go to zero or both go to infinity at some point

AND If limit f'/ g' exists then .....

We cannot make the conclusion here because the conditions of the hypothesis are not met.

Lim [x + sin(x)] / [2x – 5sin(x)]

x→∞

=

Lim [x/x + sin(x)/x] / [2x/x – 5sin(x)/x]

x→∞

=Lim [1 + sin(x)/x] / [2 – 5sin(x)/x]

x→∞

=[1 + 0] / [2 – 5*0]

=1/2

The Limit is an indeterminate form of type inf. / inf.

So L'hospitals rule can be applied once to give you..

[1 + cosx] / [2 - 5cosx]

this is not indeterminate so you can't apply L'hospital's rule again.

As x approaches infinity [1+ cosx] oscillates between 0 and 2

and [2 - 5cosx] oscillates between -3 and 7

It appears to me that this limit does not exist, but I have been wrong before.

*EDIT*

Ksoileau has proven it to be 1/2

I agree with his answer

use LH's rule when lim-> 0/0, infinity/infinity, inf./0. in this case i would say yes as limit -> inf/inf