Not L'Hopital rule

Joyce · #1 $Old$ October 5th, 2007, 08:18 AM

limx->infinite ((9X+1)^1/2)/((X+1)^1/2)

limx->90'- secX/tanx

what it means by putting positive and negative behind?how to solve by using other way?

earboth · #2 $Old$ October 5th, 2007, 08:50 AM

Quote:

Originally Posted by Joyce $View Post$

limx->infinite ((9X+1)^1/2)/((X+1)^1/2)

...

Hi,

$\lim_{x \to \infty}{\frac{\sqrt{9x+1}}{\sqrt{x+1}}} = \lim_{x \to \infty}{\frac{3 \sqrt{x} \cdot \overbrace{\sqrt{1+\frac1{9x}}}^{\text{limit is 1}}}{\sqrt{x} \cdot \underbrace{\sqrt{1+\frac1x}}_{\text{limit is 1}}}}$ = 3

earboth · #3 $Old$ October 5th, 2007, 11:08 AM

Quote:

Originally Posted by Joyce $View Post$

...

limx->90'- secX/tanx

what it means by putting positive and negative behind?how to solve by using other way?

Hi,

$\lim_{x \to \frac {\pi}{2}}\left(-\frac{\sec(x)}{\tan(x)}\right)=\lim_{x \to \frac {\pi}2}\left(-\frac{\frac{1}{\cos(x)}}{\frac{\sin(x)}{\cos(x)}}\right)=$ $\lim_{x \to \frac {\pi}2}\left(-\frac{1}{\cos(x)} \cdot \frac{\cos(x)}{\sin(x)}\right)=\lim_{x \to \frac {\pi}2}\left(-\frac{1}{\sin(x)}\right) =$ $-1$

Quote:

Originally Posted by Joyce $View Post$

what it means by putting positive and negative behind?how to solve by using other way?

I only can guess because I don't know this way of writing: Probably the - or + sign indicates the direction of approach. So 90°- could mean that you approaches the 90° from the left where the values are smaller than 90°. Ask your teacher or try to find an explanation in your textbook.

Soroban · #4 $Old$ October 5th, 2007, 11:29 AM

Hello, Joyce!

Another approach . . .

Quote:

$\lim_{x\to\infty}\frac{\sqrt{9x+1}}{\sqrt{x+2}}$

We have: . $\lim_{x\to\infty}\sqrt{\frac{9x+1}{x+2}}$

Divide top and bottom by $x\!:\;\;\lim_{x\to\infty}\sqrt{\frac{9 + \frac{1}{x}}{1 + \frac{2}{x}}} \;=\;\sqrt{\frac{9+0}{1+0}} \:=\:\sqrt{9} \;=\;3$