Question

Hello I have got problem with limit

lim x->-1(right limit)
-2*x*exp(1/(x^2-1)) / (x^2-1)^2
I know about Lhospital rule but as I use it I can not get the right result. Does anybody have any idea how to solve limits like that?
Thank you for your answer.

Answer 1

Use L hospitals rule only it is the best see if both numerator and denominator either tend to 0 or infinity then the limit will not change by differentiating both num and den.

Answer 2

I would think this answer is 0 only. How many ever times you differentiate you will still have exponential so it is 0 .

Answer 3

How can you see that so quickly? Or how can I show that?

Answer 4

Well actually if you have e^ some function tending to -infinty in the numerator then in all probability you will get 0.

Answer 5

So I do not need to differentiate it by lohospital rule.

Answer 6

Well actually in practice it is better to differentiate and see for yourself,
In this differentiating once won't do you need to differentiate 4 times so i think it would be better if you did apply lopital's rule. But the answer is still 0. Its just a shortcut observation

Answer 7

I started with lohospital like this -2*lim x->-1 x/(x^2-1)^2 / 1/exp(1/x^2-1) and get Inf/Inf
and mahe two differentiate. Is this start correct?

Answer 8

Why are you making it more complicated just differentiate the original funtion only but yeah your equation isn't wrong you can proceed i would think it would be simpler to use Lopital's rule on the initial funtion itself.

Answer 9

But when I differentiate two times I get this
- lim x-> -1 (3*x^4+6*x^2-1)*exp(1/x^2-1) /2*x^3 and then I have got no possibility to use Lhospital again.

Answer 10

Exactly so the answer is 0 when you dont get a 0/0 or infinity/infinity form then just substitue x value in the limit. As i said you still have E^ thing which is still 0 as x=-1 hence answer is 0

Answer 11

So when I insert -1 inside limit I get 0 and that is it?

Answer 12

Thanky you very much !!!

Answer 13

No problem. :)