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?
Hi,
$\displaystyle \lim_{x \to \frac {\pi}{2}}\left(-\frac{\sec(x)}{\tan(x)}\right)=\lim_{x \to \frac {\pi}2}\left(-\frac{\frac{1}{\cos(x)}}{\frac{\sin(x)}{\cos(x)}}\ right)=$ $\displaystyle \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) = $ $\displaystyle -1$
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.
Hello, Joyce!
Another approach . . .
$\displaystyle \lim_{x\to\infty}\frac{\sqrt{9x+1}}{\sqrt{x+2}} $
We have: .$\displaystyle \lim_{x\to\infty}\sqrt{\frac{9x+1}{x+2}} $
Divide top and bottom by $\displaystyle x\!:\;\;\lim_{x\to\infty}\sqrt{\frac{9 + \frac{1}{x}}{1 + \frac{2}{x}}} \;=\;\sqrt{\frac{9+0}{1+0}} \:=\:\sqrt{9} \;=\;3$