Question

find the limit x^(-17/lnx)
x->0+

Answer 1

Try taking the log

Answer 2

\[x ^{-17/lnx} x->0^{+}\]

Answer 3

taking the log of lnx, or the whole equation? I don't know if this makes a difference, but were using Hopitals rule

Answer 4

you can just take the log of the whole equation. ln(x^(-17/ln(x))=-17.

Answer 5

thus the equation is equal to the constant e^(-17).

Answer 6

thus the limit is also e^(-17)

Answer 7

the derivative of lnx is (1/x)
how were you able to go from this equation to base e. It shows the same steps in my book, but I am just a little confused to how you got there

Answer 8

(thank you so much for your help by the way)

Answer 9

1) ln(a^b)=b*ln(a)
2) ln(e)=1
3) e^ln(x)=x

Answer 10

so if ln(x^(-17/lnx))=-17, then x^(-17/lnx)=e^(-17)

Answer 11

you literally explained it so simply that my professor can't even sum it up that way

Answer 12

:)

Answer 13

i really appreciate this explanation, thank you so much!