Hey guys. Would appreciate a little help woth the following problem. Im sure its quite straight forward but im not sure on how to approach it. Well here it is:

lim $\displaystyle

\frac{e^{2x}-2e^x+1}{x^2}

$

x->0

Thanks a lot for any help.

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- Mar 8th 2008, 07:28 AMledbymischiefLimit continuity
Hey guys. Would appreciate a little help woth the following problem. Im sure its quite straight forward but im not sure on how to approach it. Well here it is:

lim $\displaystyle

\frac{e^{2x}-2e^x+1}{x^2}

$

x->0

Thanks a lot for any help. - Mar 8th 2008, 07:52 AMPaulRS
Note that: $\displaystyle (e^{2x}-2e^x+1)=(e^x-1)^2$

Thus: $\displaystyle \frac{e^{2x}-2e^x+1}{x^2}=\left(\frac{e^x-1}{x}\right)^2$

Now remember that: $\displaystyle \lim_{u\rightarrow{0}}\frac{e^u-1}{u}=1$

So, by the continuity of $\displaystyle g(x)=x^2$ we have that $\displaystyle \lim_{x\rightarrow{0}}\frac{e^{2x}-2e^x+1}{x^2}=\lim_{x\rightarrow{0}}\left(\frac{e^x-1}{x}\right)^2=\left(\lim_{x\rightarrow{0}}\frac{e ^x-1}{x}\right)^2=1$ - Mar 8th 2008, 07:56 AMledbymischief
Thanks Paul. I can't believe i missed that. Great help!