Need help with this limit!

**xxlvh** · November 8th 2009, 09:20 PM

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

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

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

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

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

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

And I don't know what to do next.

**Jhevon** · November 8th 2009, 09:27 PM

Originally Posted by xxlvh

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

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

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

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

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

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

And I don't know what to do next.

note that $\frac 1{\cot x} = \tan x$ . The derivative is $\sec^2 x$ (if you simplified your denominator, you would have gotten the same thing). Now just plug in $x = 0$ and continue

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

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