Question

can someone solve this please!

using L'Hospitals rule

(x-ln(x)) as x approaches infinity

thank you

Answer 1

x grow a lot faster than ln(x) so the limit is +Infinity

Answer 2

Can't apply L'H directly since the difference in an indeterminate form. There is What if you study the ratio of
\[\frac{x}{\ln x}\]
If this approaches infinity as x approaches infinity then you know that x grows faster, so the original problem has a limit of infinity as well.

Answer 3

You can apply L'Hospital's rule by noticing
\[
\frac{(x-\ln (x)) (x+\ln (x))}{x+\ln
(x)}=\frac{x^2-\ln ^2(x)}{x+\ln (x)}
\]

Answer 4

by taking the derivative of up and down, you get
\[
\frac{2 x - 2 \frac{\ln (x)}{x}}
{1+ \frac 1 x}

\]
You can see that the above ratio goes to \(+\infty\)

Answer 5

ok so you multiplied the numerator and the denominator by (x+ln(x)) why did you do that?

Answer 6

to be able to use L'Hospital's rule

Answer 7

what is the requirements ba to use the L'Hospitals Rule @eliassaab

Answer 8

@KingGeorge howdo you solve this

Answer 9

the requirement to use L'Hospitals rule is that the numerator/denominator should be of the form 0/0 or infinity/infinity if u directly put the value of x

Answer 10

ok then how about your final answer should it also be infinity or zero? @hartnn

Answer 11

ok,the purpose of using L'Hospitals rule is to remove such indeterminant form.
if removed by differentiation,then final answer would be finite

Answer 12

oh so the answer of elissab earlier was correct? i just need to differentiate it more?

Answer 13

@hartnn

Answer 14

yes,the answer is correct,
u will get same answer if u diff more.......so no need

Answer 15

so wait example that was given in an exam what should be my final answer? @hartnn

Answer 16

infinity only.

Answer 17

ohh ok thanks men so i will just sub infinity in the equation?

Answer 18

if u diff it just once more(after simplifying),u will get constant in denominator..
then u substitute infi,to get final answer as infinity

Answer 19

ohh ok thanks :)

Answer 20

welcome :)

Answer 21

can you solve me one last questioon before i leave?

Answer 22

@hartnn

Answer 23

sure.

Answer 24

ok i'll make another post wait ffor it

Answer 25

ok