# L'Hopital's Rule

• Jan 14th 2010, 01:31 PM
CarDoor
L'Hopital's Rule
I need help with these two problems.

$\displaystyle \lim_{x\to0^+}$ $\displaystyle (lnx-lnsinx)$

and

$\displaystyle \lim_{x\to0^+}$ $\displaystyle \frac{1}{x}$ - $\displaystyle \frac{1}{\sqrt{x}}$
• Jan 14th 2010, 01:36 PM
Jhevon
L'Hopital's rule? I wouldn't use that here...
Quote:

Originally Posted by CarDoor
I need help with these two problems.

$\displaystyle \lim_{x\to0^+}$ $\displaystyle (lnx-lnsinx)$

Hint: lnA - lnB = ln(A/B)

Quote:

$\displaystyle \lim_{x\to0^+}$ $\displaystyle \frac{1}{x}$ - $\displaystyle \frac{1}{\sqrt{x}}$
Hint: $\displaystyle \frac 1x - \frac 1{\sqrt x} = \frac {1 - \sqrt x}x$
• Jan 14th 2010, 01:39 PM
General
Quote:

Originally Posted by CarDoor
I need help with these two problems.

$\displaystyle \lim_{x\to0^+}$ $\displaystyle (lnx-lnsinx)$

and

$\displaystyle \lim_{x\to0^+}$ $\displaystyle \frac{1}{x}$ - $\displaystyle \frac{1}{\sqrt{x}}$

$\displaystyle \lim_{x\to0^+}$ $\displaystyle (lnx-lnsinx)$
$\displaystyle =\lim_{x\to0^+}ln(\frac{x}{sinx})$
$\displaystyle =ln(\lim_{x\to0^+}\frac{x}{sinx})$
$\displaystyle =ln(1)=0$

For the another one:
$\displaystyle \frac{1}{x}-\frac{1}{\sqrt{x}} = \frac{1-\sqrt{x}}{x}$

Got it ?
• Jan 14th 2010, 02:00 PM
CarDoor
For the second one, what did you do to get that?
• Jan 14th 2010, 02:22 PM
Jhevon
Quote:

Originally Posted by CarDoor
For the second one, what did you do to get that?

combine the fractions
• Jan 14th 2010, 02:33 PM
CarDoor
Oh I see. I guess I was thinking too hard. Thanks for the help.