Thread: How to evaluate this limit?

I need to evaluate the lim x->0 tan^2(x)/x

I can use any method to do this. I know the limit is zero but I wasn't sure what method I could use to evaluate this.

It could be re-written as:
(sinx/cosx)(sinx/cosx)/x

The sin of 0 is 0 and therefore makes the whole function zero, would this be sufficient enough to evaluate it?

You could use L'Hôpital's rule to write the limit as:

Hi zzizi,

lim x->0 sin(x)/x =1,
lim x->0 tan(x)/x =1,
lim x->0 tan(x)/x *tan(x) = 1*0 = 0

Thank you both very much!

Originally Posted by zzizi

It could be re-written as:
(sinx/cosx)(sinx/cosx)/x

The sin of 0 is 0 and therefore makes the whole function zero, would this be sufficient enough to evaluate it?

Watch that last comment! Yes, sin(0) = 0, but x is in the denominator so x = 0 makes the expression indeterminate.

-Dan

Originally Posted by zzizi

I need to evaluate the lim x->0 tan^2(x)/x

I can use any method to do this. I know the limit is zero but I wasn't sure what method I could use to evaluate this.

It could be re-written as:
(sinx/cosx)(sinx/cosx)/x

The sin of 0 is 0 and therefore makes the whole function zero, would this be sufficient enough to evaluate it?

Originally Posted by topsquark

Watch that last comment! Yes, sin(0) = 0, but x is in the denominator so x = 0 makes the expression indeterminate.

-Dan

Yes, perhaps I was a bit hasty there.

Originally Posted by Prove It

Wonderful! thank you sir!