Question

lim x approaches 0 (1-x)e^x-1/(x^2)

Answer 1

\[\lim_{x \rightarrow 0}\frac{(1-x)e^{x-1}}{x^2}\]

Answer 2

is that right?

Answer 3

\[\lim_{x \rightarrow 0}\frac{(1-x)e^x-1}{x^2} \]
or \[\lim_{x \rightarrow 0}(1-x)e^x-\frac{1}{x^2}\]

Answer 4

Yes

Answer 5

the first one?

Answer 6

what happens if you plug in 0?
what do you get?

Answer 7

The one above the word or

Answer 8

0 but my answer says -1/2

Answer 9

you shouldn't get 0 when you plug in 0

Answer 10

\[\frac{(1-0)e^{0}-1}{0^2}=\frac{e^0-1}{0^2}=\frac{1-1}{0}=\frac{0}{0}\]
you get 0/0 which means you have more work to do

Answer 11

if you did have 0/something other than 0 than yeah it would be 0
but 0/0 means you have more work

Answer 12

do you know l'hospital?

Answer 13

no

Answer 14

do you know this:
\[\lim_{x \rightarrow 0} \frac{e^x-1}{x}=1\]

Answer 15

For real?

Answer 16

i'm just trying to get a feel for where you are at so i can no how to assist
l'hosptial is the only way I'm seeing right now
doesn't mean there is algebraic way though

Answer 17

or isn't

Answer 18

cool I just don't understand how -1/2 is my answer

Answer 19

use the taylor expansion for \(e^x\)

Answer 20

What's that?

Answer 21

i think i have another way

Answer 22

let u=x^2 \[\lim_{u \rightarrow 0}\frac{[(1-\sqrt{u})e^{\sqrt{u}}-1]-[(1-\sqrt{0})e^{\sqrt{0}}-1]}{u-0}\]
but I wonder if i should consider both cases sqrt(x)=u and also -sqrt(x)=u

Answer 23

\[=[(1-\sqrt{u})e^\sqrt{u}-1]'|_{u=0}\]

Answer 24

and probably need to consider both cases
because that right there is actually u approaches 0 from right

Answer 25

i was just using the definition of derivative if you are confused :p

Answer 26

Oh no wonder

Answer 27

\[\lim_{u \rightarrow 0}\frac{f(u)-f(0)}{u-0}=f'(0)\]

Answer 28

you should find both right and left derivative
though they will both be the same