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.
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.
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 $\displaystyle \lim_{x\rightarrow{0}}\frac{e^{2x}-2e^x+1}{x^2}=\lim_{x\rightarrow{0}}\left(\frac{e^x-1}{x}\right)^2=\left(\lim_{x\rightarrow{0}}\frac{e ^x-1}{x}\right)^2=1$