# Thread: Help with this limit?

1. ## Help with this limit?

lim $\displaystyle (1+2x) raised to the 3csc(2x) power$
x->0

Sorry, I couldn't get it to display correctly but its (1+2x) ^ (3csc(2x)). I have it worked out by my professor, but am confused as to what he did. The first thing he did was take the natural log, and eventually the answer was $\displaystyle e^3$. How do you reach this answer?

2. Originally Posted by Mattpd
$\displaystyle \lim_{x\rightarrow 0}(1+2x)^{3\csc(2x)}$
$\displaystyle \lim_{x\rightarrow 0}\ (1+2x)^{3\csc(2x)} = \lim_{x\rightarrow 0}\ \exp(\ln{(1+2x)^{3\csc(2x)}}) = \lim_{x\rightarrow 0}\ \exp(3 \csc(2x)\ln{(1+2x)})$ $\displaystyle = \lim_{x\rightarrow 0}\ \exp(3\frac{\ln{(1+2x)}}{\sin(2x)}) = \lim_{x\rightarrow 0}\ \exp(3 \frac{\frac{2}{1+2x}}{2\cos(2x)}) = e^{3}$

Where I used L'Hospitals in the second to last equality.

3. $\displaystyle \lim_{x \to 0} \ (1+2x)^{3 \csc(2x)}$

let $\displaystyle y = \ln \Big((1+2x)^{3 \csc(2x)}\Big) = 3 \csc (2x) \ln (1+2x) = \frac{3 \ln(1+2x)}{\sin(2x)}$

then $\displaystyle \lim_{x \to 0} y = \lim_{x \to 0} \frac{3 \ln(1+2x)}{\sin(2x)}$

now use L'Hospital's rule

$\displaystyle = \lim_{x \to 0} \ \frac{3 \ \frac{2}{1+2x}}{2 \cos(2x)} = \frac{6}{2} = 3$

so $\displaystyle \lim_{x \to 0} \ (1+2x)^{3 \csc(2x)} = \lim_{x \to 0} e^{y} = e^{\lim_{x \to 0}y} = e^{3}$

4. Hello, Mattpd!

This is a messy one . . .

$\displaystyle \lim_{x\to0}(1+2x)^{3\csc2x}$

Let: .$\displaystyle y \:=\:(1+2x)^{3\csc2x}$

Take logs: .$\displaystyle \ln(y) \;=\;\ln(1+2x)^{3\csc2x} \;=\;3\csc2x\cdot\ln(1+2x)$

We have: .$\displaystyle \ln(y) \;=\;3\,\frac{\ln(1+2x)}{\sin2x}$ . . . which goes to $\displaystyle \frac{0}{0}$

Apply L'Hopital: .$\displaystyle 3\cdot\frac{\frac{1}{1+2x}\cdot2}{2\cos2x} \;=\;\frac{3}{(1+2x)\cos2x}$

Then: .$\displaystyle \lim_{x\to0}\,\ln(y) \;=\;\lim_{x\to0}\frac{3}{(1+2x)\cos2x} \;=\;\frac{3}{1\cdot1} \;=\;3$

We have: .$\displaystyle \lim_{x\to0}\,\ln(y) \:=\:3$

Therefore: .$\displaystyle \lim_{x\to0} y \;=\;e^3$

Thanks all.