# Math Help - l'Hopital's rule....help

1. ## l'Hopital's rule....help

Find the lim as 'x' approaches infinity $(1+2x)^{(17)/(2lnx)}$ (sorry dont know how to put the lim on here....) thanks for any help..

2. Originally Posted by omibayne
Find the lim as 'x' approaches infinity $(1+2x)^{(17)/(2lnx)}$ (sorry dont know how to put the lim on here....) thanks for any help..
You have $\lim_{x\to\infty}\left(1+2x\right)^{\frac{17}{2\ln (x)}}$

Must you use L'hopital's? To make it a little simpler try letting $\ln(x)=z$ so our limit becomes $\lim_{z\to\infty}\left(1+2e^z\right)^{\frac{17}{2z }}=L\Rightarrow{L^{\frac{2}{17}}}=\lim_{z\to\infty }\left(1+2e^z\right)^{\frac{1}{z}}$

So then

$\ln\left(L^{\frac{2}{17}}\right)=\lim_{z\to\infty} \frac{\ln\left(1+2e^z\right)}{z}$

Either applying L'hopitals (or noticing that $\ln\left(1+2e^z\right)\sim{z}$) gives us

$\lim_{z\to\infty}\frac{\ln\left(1+2e^z\right)}{z}= \lim_{z\to\infty}\frac{\frac{2e^z}{1+2e^z}}{1}=1$

So if $\ln\left(L^{\frac{2}{17}}\right)=1\Rightarrow{L=e^ {\frac{17}{2}}}$

3. Originally Posted by Mathstud28
You have $\lim_{x\to\infty}\left(1+2x\right)^{\frac{17}{2\ln (x)}}$

Must you use L'hopital's? To make it a little similar try letting $\ln(x)=z$ so our limit becomes $\lim_{z\to\infty}\left(1+2e^z\right)^{\frac{17}{2z }}=L\Rightarrow{L^{\frac{2}{17}}}=\lim_{z\to\infty }\left(1+2e^z\right)^{\frac{1}{z}}$

So then

$\ln\left(L^{\frac{2}{17}}\right)=\lim_{z\to\infty} \frac{\ln\left(1+2e^z\right)}{z}$

Either applying L'hopitals (or noticing that $\ln\left(1+2e^z\right)\sim{z}$) gives us

$\lim_{z\to\infty}\frac{\ln\left(1+2e^z\right)}{z}= \lim_{z\to\infty}\frac{\frac{2e^z}{1+2e^z}}{1}=1$

So if $\ln\left(L^{\frac{2}{17}}\right)=1\Rightarrow{L=e^ {\frac{17}{2}}}$
thanks alot for ur help...