Question

evaluate limit. as limit approaches negative infinity, x to the 2 times e to the 2x

Answer 1

\[\lim_{x \rightarrow - \infty} x^{2e^{2x}}\]

Answer 2

\[\lim_{x \rightarrow -\infty}x^2e^{2x}\]

Answer 3

^^ ahh ok

Answer 4

That limit will be 1 because as x approaches - infinity, e^2x goes to 0.. 2 X 0 is 0 and x^0 is 1.

Answer 5

x^2 approaches -infinity but e^2x approaches 0. We have an indeterminant form here and need to use l'Hospital's rule.\[=\lim_{x \rightarrow -\infty}\frac{x^2}{e^{-2x}}=\lim_{x \rightarrow -\infty}\frac{2x}{-2e^{-2x}}\]We still have an indeterminant form here -ifinity/-infinity. We apply l'Hospital's rule a second time:\[=\lim_{x \rightarrow -\infty}\frac{1}{2e^{-2x}}=0\]