Finding the limit

**BACONATOR** · April 1st 2008, 10:47 PM

I've got these two problems.
Any help would be appreciated!
Thanks!

**TheEmptySet** · April 1st 2008, 10:51 PM

Originally Posted by BACONATOR

I've got these two problems.
Any help would be appreciated!
Thanks!

$\lim_{x \to \infty}x^2e^{-4x} = \frac{x^2}{e^{4x}}$

using L'hospitals rule twice we get

$\lim_{x \to \infty}\frac{2x}{4e^{4x}}=\lim_{x \to \infty}\frac{2}{16e^{4x}}=\frac{2}{\infty}=0$

**TheEmptySet** · April 1st 2008, 10:58 PM

$f(x)=\begin{cases}<br /> \frac{x^2+x-12}{x-3}=\frac{(x+4)(x-3)}{(x-3)}=(x+4) \mbox{ if } x < 3 \\<br /> ae^{x-3} \mbox{ if } x \ge 3<br /> \end{cases}$

so to be cont at x=3 we need

$(3+4)=ae^{3-3} \iff 7=a$

for part b use the limit defintion.

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