Can L’Hopital’s rule be used for this problem?

Question

Lim [x + sin(x)] / [2x – 5sin(x)]
x→∞

I know what the answer is. I just want to know if L’Hopital’s rule can be used to find that limit. If not, why not; and if so, show me how.

Steiner: So you're saying that LH rule doesn't apply because it doesn't work when you try it? If all of the required conditions are met, why doesn't it work?

Is there like a bug in LH rule?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

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