# Math Help - limit using l'hopital

1. ## limit using l'hopital

this is my question:

lim [tan(x)/(x)]/(1/(x^2)
x->0

I have tried to work this out and I know how to get to: [ ln(tan(x)/x) ]/x^2
but then when I start using l'hopital I get stuck.
PLEASE DO NOT TELL ME THAT IT'S e^1/3, because I want to see the way to solve this and not the final answer!

THANKS!

2. $\lim_{x\rightarrow{0}}\frac{\frac{tan(x)}{x}}{\fra c{1}{x^{2}}}=\lim_{x\rightarrow{0}}[xtan(x)]$

Are you sure you posted it correctly. Not much need for L'Hopital. The limit is merely 0.

3. Originally Posted by Lilly100
this is my question:

lim [tan(x)/(x)]/(1/(x^2)
x->0

I have tried to work this out and I know how to get to: [ ln(tan(x)/x) ]/x^2
but then when I start using l'hopital I get stuck.
PLEASE DO NOT TELL ME THAT IT'S e^1/3, because I want to see the way to solve this and not the final answer!

THANKS!
Originally Posted by galactus
$\lim_{x\rightarrow{0}}\frac{\frac{tan(x)}{x}}{\fra c{1}{x^{2}}}=\lim_{x\rightarrow{0}}[xtan(x)]$

Are you sure you posted it correctly. Not much need for L'Hopital. The limit is merely 0.
Not to mention that when you apply L'Hopital's rule to this problem I get that
$\lim_{x \to 0}\frac{\frac{tan(x)}{x}}{\frac{1}{x^{2}}} = \lim_{x \to 0} \frac{x(sin(x)~cos(x) - x)}{2~cos^2(x)}$
not what you wrote.

-Dan

4. ## what I posted

This is the original limit:

lim [tan(x)/(x)]/(1/(x^2)
x->0

tan(x)/(x) is the base
1/(x^2) is the power

galactus:
notice that tanx/x is in ln form

5. It would've helped if you would have told us 1/x^2 was a power in the first place.

Sorry for the small font. I am unable to make it larger.

Rewrite as:

$e^{\lim_{x\rightarrow{0}}\frac{ln(\frac{tan(x)}{x} )}{x^{2}}}$

Using L'Hopital:

$e^{\frac{1}{2}\lim_{x\rightarrow{0}}\frac{x+xtan^{ 2}x-tanx}{x^{2}tanx}}$

L'Hopital again:

$e^{\lim_{x\rightarrow{0}}\frac{tanx}{2tanx+x+xtan^ {2}x}\cdot\lim_{x\rightarrow{0}}[1+tan^{2}x]}$

The limit to the right is just 1 so we can shed it and use L'Hopital again:

$e^{\lim_{x\rightarrow{0}}\frac{1}{2xtanx+3}}$

Of course, after all this we should be able to see the limit is 1/3 and we have:

$e^{\frac{1}{3}}$

You could also break it up and attempt another route.

$\frac{tan(x)}{x}=\frac{1}{cos(x)}\cdot\frac{sin(x) }{x}$

$e^{\frac{ln(\frac{1}{cos(x)})\cdot\frac{sin(x)}{x} )}{x^{2}}}$

$e^{\frac{ln(sec(x))}{x^{2}}}\cdot{e^{\frac{ln(\fra c{sin(x)}{x})}{x^{2}}}}$

The left limit is $\sqrt{e}$ and the right one is $e^{\frac{-1}{6}}$

$e^{\frac{1}{2}}\cdot{e^{\frac{-1}{6}}}=e^{\frac{1}{3}}$

There are many ways to go about it. The method WITHOUT L'Hopital is the most interesting.

6. galactus!

Thank you for solving this.

If you could explain the first l'hopital that you did, that would be great!

when I did l'hopital for this-

ln[tan(x)/(x)]/x^2

i got to:
for the numerator {1/ [tan(x)/(x)]*[1/(cosx)^2]*x-[tan(x)*1]} / x^2
for the denominator 2x

what am I doing wrong?

thanks again

7. I just took the derivative of $ln(\frac{tan(x)}{x})$

Which is $\frac{x+xtan^{2}(x)-tan(x)}{x^{2}tan(x)}$

Chain rule:

Derivative of outside:

$\frac{1}{\frac{tan(x)}{x}}=\frac{x}{tan(x)}$

Derivative of inside:

$\frac{d}{dx}[\frac{tan(x)}{x}]=\frac{x(1+tan^{2}(x))-tan(x)}{x^{2}}$

$\frac{x}{tan(x)}\cdot\frac{x(1+tan^{2}(x))-tan(x)}{x^{2}}$

= $\frac{x+xtan^{2}(x)-tan(x)}{xtan(x)}$