Question

how to use L'Hopital's rule on this problem?
lim x -> ∞ (x + e^x)^1/x

Answer 1

I just learned L'Hopital myself, so I might not be able to help, but do you know how to start ?

Answer 2

not at all =/

Answer 3

Hi
\[\lim_{x \rightarrow \infty}(x+e^{x})^{1/x} \rightarrow \lim_{x \rightarrow \infty}1/x*\ln(x+e^x)\]
\[hopital-->\lim (1+e^x)/(x+e^x)-->hopital-->e^x/e^x=1 \]
\[answer=e^1=e\]

Answer 4

thanks alot but i dont get much

Answer 5

You need to see if the limit is of the inderterminant form
\[\frac{ 0 }{ 0 }, or \frac{ \pm \infty }{ \pm \infty }\]
If it is, then you can apply L'Hopital

Answer 6

or \[\infty^0\] which is the case here

Answer 7

ok

Answer 8

so, to bring down the exponent you have to take the ln of both sides (to use the property of logarithm)

Answer 9

oh ths

Answer 10

when you use ln the expression becames \[\infty/\infty\]
in the last step \[e^{\ln x}=x\]

Answer 11

ok

Answer 12

\[\ln y = \lim_{x \rightarrow \infty} (\frac{ 1 }{ x } \ln (1+e^x))\]
You still have the form
\[\frac{ \infty }{ \infty }\]
so now you can apply L'Hopital

Answer 13

ok - you got your answer

Answer 14

ok ths

Answer 15

do you get how to solve it ?

Answer 16

i thnik so

Answer 17

i need to write it down step by sstep maybe i will get it all

Answer 18

wut do i do when i bring down the exponent again

Answer 19

i got 1/x * ln(x+e^x)

Answer 20

what is the next step after you get
\[\ln (y) = \lim_{x \rightarrow \infty}\left[ \frac{ \ln(1+e^x) }{ x } \right]\]

Answer 21

right, now you brought down the exponent, o now you have the inderterminant form infinity/infinity
So now you can apply L'Hopital, meaning take the derivative of the numerator and the derivative of the denominator

Answer 22

ok

Answer 23

i got ln(y) = e^x / x + e^x

Answer 24

no, check your denominator again

Answer 25

i got x = 1

Answer 26

yes, so you get (1 + e^x)(1) = 1 + e^x

Answer 27

so now you have (e^x)/(1+e^x), that is the inderterminant form of infinity / infinity, so you can apply L'Hopital again

Answer 28

i took the derivative and got e^x / x + e^x because the derivate of ln is 1/x ?

Answer 29

let me double check my answer, I think I got it right

Answer 30

ok

Answer 31

\[\frac{ d }{ dx } \ln(1 + e^x) = \frac{ \frac{ d }{ dx } (1 + e^x) }{ (1 + e^x) } = \frac{ e^x }{ 1 + e^x }\]
Isn't that correct ?

Answer 32

yes

Answer 33

and d/dx if x = 1, so you have the denominator (1+e^x)(1), where (1) is the derivative of x

Answer 34

so
\[\ln (y) = \lim_{x \rightarrow \infty}\left[ \frac{ e^x }{ (1+e^x)(1) } \right] = \frac{ \infty }{ \infty }\]

Answer 35

ohh ok

Answer 36

I think the problem was x+e^x so you have to write
\[d/dx \ln(x+e^x)=(1+e^x)/(x+e^x)\]

Answer 37

and if you apply L'Hopital again, then you get e^x / e^x which cancels and become 1
Then
ln(y) = 1

Answer 38

ohhhh - sorry - I missed that, I was doing 1 + e^x

Answer 39

ok

Answer 40

thanks for walking me through, i think i got it

Answer 41

see, I use (1+..) every where , ok - my mistake, I will go back and redo it :)
But the steps should still be the same, if you get the inderterminant form infinity/infinity then apply L'hopital again,
at the end you get
ln(y) = 'something'
then to find y you raise both sides by e

Answer 42

ok

Answer 43

ths