Question

Fast limit:

Answer 1

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

Answer 2

I have no idea o.o'

Sorry :/ Wish I c00d help.

Answer 3

I have an idea but I'm stuck:
\[L'HOPITAL:\left[ \frac{ 0 }{ 0} \right]=>\lim_{x \rightarrow 0^{-}}\frac{ \frac{ -e^\frac{ 1 }{ x } }{ x^2 } }{ 2x }\]
It's getting worse to a nonsense!

Answer 4

@AravindG

Answer 5

How did you get 0/0 form? e^infinity=infinity

Answer 6

e^(-∞)=0
ln(0)=-∞

Answer 7

Looks like nobody's into calculus on this platform

Answer 8

No, there are people into calculus here.

Answer 9

@phi

Answer 10

\[
\frac{e^{1/x}}{x^2}= \exp\left(\ln\left(\frac{e^{1/x}}{x^2}\right)\right) = \exp\left(\frac 1x-2\ln(x)\right)
\]

Answer 11

This holds due to \(x>0\)

Answer 12

x->0(-)=>x<0;
For x>0 case the limit is simple is +∞;

Answer 13

Wait, what I should have said is that \[
\frac{e^{1/x}}{x^2}>0
\]

Answer 14

And if you are worried about the sign, then keep it as: \[
\ln(x^2)
\]

Answer 15

Or make it \(2\ln|x|\)

Answer 16

wio still not helping we're at limits

Answer 17

ya it is

Answer 18

\[
\lim_{x\to0^-}\exp\left(\frac 1x - 2\ln|x|\right)
\]

Answer 19

Finally I got it thanks wio