Question

Would I use L'Hospitals rule to solve

Lim
x-->INF

x^2e^-x

Answer 1

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

Answer 2

Yeah, but you'd need to apply it more than once.

Answer 3

thank you

Answer 4

yep

Answer 5

so would \[\lim_{x \rightarrow \infty} x^2e ^{-x} = 0\]

Answer 6

you're really looking at \(\large \sf \frac{x^2}{e^x}\)
if that helps.

Answer 7

how I got that is first I rewrote the problem as \[\lim_{x \rightarrow \infty} \frac{ x^2 }{ e^x }\] and that would be \[\infty \over \infty \]

Answer 8

so do it again

\[\lim_{x \rightarrow \infty} \frac{ 2x }{ e^x } \]
is \[\infty \over \infty \]

Answer 9

do it again.

Answer 10

L'hopitals can be applied more than once, and often is the case.

Answer 11

then do hospitals again

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

limit is now \[2 \over \infty \] which is like saying 0

Answer 12

\[\sf \lim_{x \rightarrow ∞} \frac{1}{e^x} \]

Answer 13

which is something you should know by now.

Answer 14

why would it be 1 over? the d/dx of
(2x) is 2?

Answer 15

d\dx of 2x = 2
factor out
\[\sf 2 \times \lim_{x \rightarrow ∞} \frac{1}{e^x}\]

Answer 16

"factor"

Answer 17

ok thank you