Limit continuity

**ledbymischief** · March 8th 2008, 07:28 AM

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 $<br /> \frac{e^{2x}-2e^x+1}{x^2}<br />$
x->0

Thanks a lot for any help.

**PaulRS** · March 8th 2008, 07:52 AM

Note that: $(e^{2x}-2e^x+1)=(e^x-1)^2$

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

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

So, by the continuity of $g(x)=x^2$ we have that $\lim_{x\rightarrow{0}}\frac{e^{2x}-2e^x+1}{x^2}=\lim_{x\rightarrow{0}}\left(\frac{e^x-1}{x}\right)^2=\left(\lim_{x\rightarrow{0}}\frac{e ^x-1}{x}\right)^2=1$

**ledbymischief** · March 8th 2008, 07:56 AM

Thanks Paul. I can't believe i missed that. Great help!

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