# limit using l'hopital

• Dec 22nd 2007, 06:43 AM
Lilly100
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!
• Dec 22nd 2007, 08:36 AM
galactus
$\displaystyle \lim_{x\rightarrow{0}}\frac{\frac{tan(x)}{x}}{\fra c{1}{x^{2}}}=\lim_{x\rightarrow{0}}[xtan(x)]$

Are you sure you posted it correctly. Not much need for L'Hopital. The limit is merely 0.
• Dec 22nd 2007, 09:01 AM
topsquark
Quote:

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!

Quote:

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

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
$\displaystyle \lim_{x \to 0}\frac{\frac{tan(x)}{x}}{\frac{1}{x^{2}}} = \lim_{x \to 0} \frac{x(sin(x)~cos(x) - x)}{2~cos^2(x)}$
not what you wrote.

-Dan
• Dec 22nd 2007, 12:11 PM
Lilly100
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
• Dec 22nd 2007, 12:51 PM
galactus
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:

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

Using L'Hopital:

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

L'Hopital again:

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

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

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

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

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

You could also break it up and attempt another route.

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

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

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

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

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

There are many ways to go about it. The method WITHOUT L'Hopital is the most interesting.
• Dec 24th 2007, 03:41 AM
Lilly100
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
• Dec 24th 2007, 09:50 AM
galactus
I just took the derivative of $\displaystyle ln(\frac{tan(x)}{x})$

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

Chain rule:

Derivative of outside:

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

Derivative of inside:

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

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

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