Question

limit x-> infinity (e^3x-5x)^1/x) please help me on this... i got infinity over infinity so i used l'hopitals rule... but i couldn't complete the solution... my friend said its e^3, but i dont know how to get it...

Answer 1

Just to be clear, is it \(\Large\displaystyle \lim_{x \rightarrow\infty} (e^{3x}-5x)^{\frac{1}{x}}\)?

Answer 2

yes...

Answer 3

my solution is up to..

ln y = lim x-> infinity (3e^3x - 5)/(e^3x - 5x) .. what will i do next?

Answer 4

So in it's current state, it's approaching the indeterminate form: \(\Large\rm \infty^0\)
So we uhh, rewrite our expression using log and exponential,\[\LARGE e^{\ln\lim_{x \rightarrow\infty} (e^{3x}-5x)^{\frac{1}{x}}}\]Ignore the e for now, applying rules of logs,\[\Large\rm \lim_{x\to\infty}\frac{1}{x}\ln(e^{3x}-5x)\]Then,\[\Large\rm \lim_{x\to\infty}\frac{\ln(e^{3x}-5x)}{x}\]Now it's approaching \(\Large\rm \dfrac{\infty}{\infty}\)So this is where you applied L'Hopital I assume?

Answer 5

\[\Large\rm \lim_{x\to\infty}\frac{3e^{3x}-5}{e^{3x}-5x}\]Woops.

Answer 6

@zepdrix yeah...

Answer 7

So we're STILL getting the indeterminate form \(\Large\rm \dfrac{\infty}{\infty}\), yes?
So L'Hop again! :)

Answer 8

i got it! thanks for the info :)

Answer 9

cool c: just had to L'Hop a ton of times for this one lol

Answer 10

what will be the limit, if i use the same function, as x->0...?

Answer 11

Hmm well let's see.. everything would be the same up to this point:\[\Large\rm \lim_{x\to\infty}\frac{3e^{3x}-5}{e^{3x}-5x}\]Before this point you would be getting the indeterminate form 0/0 allowing you to apply L'Hop as you had before.

As x->0, this expression is approaching:\[\Large\rm \lim_{x\to\infty}\frac{3e^{3x}-5}{e^{3x}-5x}=\frac{3(1)-5}{(1)-0}\]The exponentials are approaching 1 as x approaches 0, yah?

So looks like we get uhhhhhhh, e^{-2} maybe?
I dunno I'll try to confirm it on Wolfram or something.

Answer 12

Wolfram agrees! Yay!

Answer 13

Ah I didn't change the subscript to x->0 on the limits! My bad.

Answer 14

thanks :)