Question

lim (x^3) (e^-x^2)
x->infin

Answer 1

... are you allowed to use L'hospital's Rule ?

Answer 2

\[\lim_{x \rightarrow \infty} x ^{3} e ^{-x ^{2}}\]

Answer 3

yes

Answer 4

\[\huge \lim_{x\to \infty }x^3 \cdot e^{(-x)^2}\]

http://www.wolframalpha.com/input/?i=lim_%7Bx%5Cto+%5Cinfty%7D+%28x%5E3%29+%28e%5E-x%5E2%29

Answer 5

no, neg outside paren

Answer 6

Ehh.. yes I made a mistake , anyway the link is ok. check it out

Answer 7

you have to use L'hospital's Rule twice ...

Answer 8

\[\LARGE \lim_{x\to\infty}x^3\cdot e^{-x^2}=\lim_{x\to \infty} \frac{x^3}{e^{x^2}} \] because ... \[\LARGE x^{-a}=\frac{1}{x^a}\]

Answer 9

k, got that part

Answer 10

.. is there anything you don't understand?

Answer 11

deriv of denom?

Answer 12

\[\LARGE (e^u)'=e^u\cdot u'\] ??

Answer 13

\[\LARGE \left(e^{x^2}\right)'=e^{x^2}\cdot (x^2)'=e^{x^2}\cdot 2x\]

Answer 14

\[\LARGE (x^2)'=2\cdot x^{2-1}=2x\]

Answer 15

ok, then that gives me inf/inf .... now how to take deriv of that denom?

Answer 16

which part do you mean? ... you have to use L'hospital's Rule twice, I told you before.

Answer 17

the 2xe^x^2 part

Answer 18

\[\LARGE \frac32 \left(\lim_{x\to\infty} \;\;\frac{1}{2x\cdot e^{x^2}}\right)\] this one ?

Answer 19

\[\lim_{x \rightarrow \infty}\frac{ 3x ^{2}}{2xe ^{x ^{2}}}\]

Answer 20

so if i plug in infinity i get inf/inf. so have to take L'H Rule again....

Answer 21

that's where i'm lost

Answer 22

Click on show steps on
http://www.wolframalpha.com/input/?i=lim_%7Bx%5Cto+%5Cinfty%7D+%28x%5E3%29+%28e%5E-x%5E2%29

Answer 23

\[\LARGE \lim_{x\to \infty}\frac{3x^2}{2x\cdot e^{x^2}}=\lim_{x\to \infty}\frac{3x}{2\cdot e^{x^2}}=\]

\[\LARGE \frac32 \cdot \left(\lim_{x\to \infty}\frac{x}{ e^{x^2}}\right)\]

note that if we substitute we get \[\LARGE \frac{\infty}{\infty}\] so we have to use L'H rule again !...

\[\LARGE =\frac32 \cdot \left[\lim_{x\to \infty}\frac{x'}{( e^{x^2})'}\right]=\]

\[\LARGE =\frac32 \cdot \left[\lim_{x\to \infty}\frac{1}{(x^2)' \cdot e^{x^2}}\right]=\]

\[\LARGE =\frac32 \cdot \left[\lim_{x\to \infty}\frac{1}{2x \cdot e^{x^2}}\right]=\frac32\cdot \frac12 \cdot \left[\lim_{x\to \infty}\frac{1}{x \cdot e^{x^2}}\right]=\]

\[\LARGE =\frac34 \cdot \left[\lim_{x\to \infty}\frac{1}{x \cdot e^{x^2}}\right]\] substituting we get...

\[\LARGE =\frac34\cdot \frac{1}{\infty}=\frac{3}{\infty}=0\] does this help ?

Answer 24

;)

Answer 25

yes, perfect. i finally got it. thank you.

Answer 26

My preasure xD