# Thread: Calculation of a limit using L'Hospital

1. ## Calculation of a limit using L'Hospital

Hello

I'm having trouble calculating the following limit:

limit as x goes to zero of ( x^2 * sin(1/x) ) / sinx

I can see that it's an indeterminate form of 0/0 so it seems like classic L'Hospital,
but when I take the limit of the derivatives it comes out:

(2x * sin(1/x) - cos(1/x) ) / cosx

and cos(1/x) has no limit at zero, so I don't know what to do.

2. The problem is that you can't be sure that x^2 sin(1/x) goes to 0, since sin(1/x) does not have a limit at 0... So you can't use L'Hospital.

3. I see...so could you use the sandwich theorem, since

(x^2 * -1) / sinx <= (x^2 * sin(1/x)) / sinx <= (x^2 * 1) / sinx

and show with L'Hopital that the limits of the leftmost side and the rightmost side at zero are zero?

4. Yes that would work.

5. Originally Posted by moses
Hello

I'm having trouble calculating the following limit:

limit as x goes to zero of ( x^2 * sin(1/x) ) / sinx

I can see that it's an indeterminate form of 0/0 so it seems like classic L'Hospital,
but when I take the limit of the derivatives it comes out:

(2x * sin(1/x) - cos(1/x) ) / cosx

and cos(1/x) has no limit at zero, so I don't know what to do.
This is [probably...] one of the [unusual...] cases where direct application of l'Hopital rule doesn't work. Anyway You can avoid the 'trap' considering that is...

(1)

... and then apply l'Hopital rule and find the value of the limit...

(2)

Kind regards

$\chi$ $\sigma$