Question

Calculate the limit :

lim (x->0-) ln(e^x)/(e^1/x)

Answer 1

\[\lim_{x \rightarrow 0^{-}} x^2/e = 0\]

Answer 2

since\[ \ln (e^x) = x\]

Answer 3

No, it's e^(1/x) not (e^1)/x ... my mistake...

Answer 4

\[-\infty\]

Answer 5

\[\frac{lne^\frac{1}{x}}{e^\frac{1}{x}}?\]

Answer 6

Imagine it as so : if f(x) = e^x , then calculate : lim (x->0) lnf(x)/f(1/x)

Answer 7

The solution is -inf but I need a more detailed answer please

Answer 8

\[\frac{\ln(e^x)}{e^\frac{1}{x}}\]?

Answer 9

Yes myininaya

Answer 10

you need to use l'hospital's rule since you get a 0/0 indeterminate

Answer 11

first you can simplify first to lim x-> 0- x/e^(1/x)

Answer 12

then the derivative of the numerator is 1

Answer 13

and the derivative of the denominator is e^(1/x) lnx

Answer 14

Believe it or not this was in my high school exams and it gave you 5/100 points... Greece rocks...

Answer 15

so you should get as a result lim x->0- 1/[e^(1/x) lnx]

Answer 16

o whoops

Answer 17

derivative of the denominator should be -1/x^2

Answer 18

If you do that it keeps getting more complicated. I tried...

Answer 19

the limit x->0- of e^1/x is 0

Answer 20

that goes to infinity not zero kanade

Answer 21

and theres a theorem that says that when lim x->a 1/0 = inf

Answer 22

your coming in from the left not the right

Answer 23

if you come in from the right it is infinity

Answer 24

and so since u took l'hospitals once it is lim x->-0 -x^2/e^(1/x)

Answer 25

but you need to remember that it is not exactly zero, it's approaching zero from the left

Answer 26

ok you are right kadeo it is zero im sorry

Answer 27

i had earlier predicted that!!!

Answer 28

im pretty sure u'll yeild different results if you approach the same equation from the right

Answer 29

since the function isnt continuous

Answer 30

Actually if you make it lim(x->0-) e^(-1/x) . ln(e^x) it gets a bit easier.

Answer 31

why do you keep writing it as ln(e^x) instead of x?
ln(e^x)=xlne=x(1)=x

Answer 32

i just feel like im missing something or im misreading what you meant to say or said

Answer 33

yea the ln(e^x) doesnt seem to hold much meaning...

Answer 34

if it can just be written as x

Answer 35

Yes, x e^(-1/x)

Answer 36

OK it seems like nobody got it right, so here's the solution : lim (x->0-) lne^x/e^(1/x) = lim x . e(-1/x) = lim e(-1/x) / (1/x) = (De L'Hospital) = lim e(-1/x) . (1/x^2)/(-1/x^2) = - lim e (-1/x) = -inf