# Thread: Need help with this limit!

1. ## Need help with this limit!

$\displaystyle \lim_{x\to0^+}(9x+1)^{cotx}$

Hopefull I am at least on the right track, here's what I tried to do:

Let $\displaystyle y=(9x+1)^{cotx}$

$\displaystyle \lim_{x\to0^+}lny=cot(x)ln(9x+1)$

$\displaystyle \lim_{x\to0^+}lny=\frac{ln(9x+1)}{\frac{1}{cot(x)} }$

$\displaystyle \lim_{x\to0^+}lny=\frac{\frac{9}{9x+1}}{\frac{csc^ 2x}{cot^2x}}$

And I don't know what to do next.

2. Originally Posted by xxlvh
$\displaystyle \lim_{x\to0^+}(9x+1)^{cotx}$

Hopefull I am at least on the right track, here's what I tried to do:

Let $\displaystyle y=(9x+1)^{cotx}$

$\displaystyle \lim_{x\to0^+}lny=cot(x)ln(9x+1)$

$\displaystyle \lim_{x\to0^+}lny=\frac{ln(9x+1)}{\frac{1}{cot(x)} }$

$\displaystyle \lim_{x\to0^+}lny=\frac{\frac{9}{9x+1}}{\frac{csc^ 2x}{cot^2x}}$

And I don't know what to do next.
note that $\displaystyle \frac 1{\cot x} = \tan x$. The derivative is $\displaystyle \sec^2 x$ (if you simplified your denominator, you would have gotten the same thing). Now just plug in $\displaystyle x = 0$ and continue

(You should have the $\displaystyle \lim_{x \to 0^+}$ on the right side of the equation also)