Question

L'hopitals rule, help

Answer 1

\[\lim_{x \rightarrow oo}15xe^{\frac{ 1 }{ x }}-15x\]

Answer 2

i already know the answer is 15

Answer 3

but dont know how to get it

Answer 4

i guess factoring out the \(15x\) doesn't help much

Answer 5

i ended up rewriting it in f(x)/g(x), by multiplying both sides by the conjugate.

Answer 6

not sure where it goes from there, or if thats the right step

Answer 7

you are going to have to write it as a fraction somehow
does taking the log help?

Answer 8

not sure can i take the log of num and denom?

Answer 9

i feel like that would change the value of it

Answer 10

you would have to exponentiate at the end, if that is a word
how about multiplying top and bottom by \(e^{\frac{1}{x}}+1\)?

Answer 11

no that does not work either damn

Answer 12

how to take the derivative of e^(1/x)

Answer 13

that i can do

Answer 14

\[-\frac{e^{\frac{1}{x}}}{x^2}\] by the chain rule

Answer 15

so if its e^(2/x) it would just be 2 times that?

Answer 16

it would be
\[e^{\frac{2}{x}}\times \frac{d}{dx}[\frac{2}{x}]=-\frac{2e^{\frac{2}{x}}}{x^2}\]

Answer 17

this would be real easy if we could use power series

Answer 18

teacher never went over that...damn, not sure here

Answer 19

wondering if theres some kind of algebra trick here, the answer is 15 according to wolfram and it came up as correct on my online hwk so.

Answer 20

yea but i don't even know what form this is in
ignore the 15
this looks like
\[\infty\times 0-\infty\]

Answer 21

if you write it as
\[x(e^{\frac{1}{x}}-1)\] then it looks like \(\infty\times -1\) very strange

Answer 22

@TylerD i am stuck too
been trying to figure it out, so i posted it myself
take a look