# Math Help - How to evaluate this limit?

1. ## 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?

2. ## Re: How to evaluate this limit?

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

$2\lim_{x\to0}\tan(x)\sec^2(x)=0$

3. ## Re: How to evaluate this limit?

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

4. ## Re: How to evaluate this limit?

Thank you both very much!

5. ## Re: How to evaluate this limit?

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

6. ## Re: How to evaluate this limit?

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?
\displaystyle \begin{align*} \lim_{ x \to 0}\frac{\tan^2{x}}{x} &= \lim_{x \to 0}\frac{\sin{x}}{x} \cdot \lim_{x \to 0} \frac{\sin{x}}{\cos^2{x}} \\ &= 1 \cdot \frac{0}{1^2} \\ &= 0 \end{align*}

7. ## Re: How to evaluate this limit?

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.

8. ## Re: How to evaluate this limit?

Originally Posted by Prove It
\displaystyle \begin{align*} \lim_{ x \to 0}\frac{\tan^2{x}}{x} &= \lim_{x \to 0}\frac{\sin{x}}{x} \cdot \lim_{x \to 0} \frac{\sin{x}}{\cos^2{x}} \\ &= 1 \cdot \frac{0}{1^2} \\ &= 0 \end{align*}
Wonderful! thank you sir!