Question

Find the indicated limit
lim->infinity (e^2x - 1)/(x)

Answer 1

use L'Hopital's rule...

Answer 2

how do i do that?

Answer 3

page 2, second column...
http://tutorial.math.lamar.edu/pdf/Calculus_Cheat_Sheet_All.pdf

Answer 4

take the derivative of (e^2x-1) and (x). it ends (2e^2x)/1

Answer 5

and then what? @sumeer

Answer 6

infinite i think :)

Answer 7

and then what? @sumeer

Answer 8

the result of question is positive infinite

Answer 9

\[
L=\lim_{x\to\infty}\frac{e^{2x}-1}{x}
\]
and positive infinity or negative infinity?

Answer 10

yes the answer for \(x\to+\infty\) is \(\infty\). he function is a continuously increasing function
but,
for \(x\to-\infty\), the answer would be "0"

Answer 11

positive infinity

Answer 12

did you study L'hopital's rules?

Answer 13

yea

Answer 14

so, would it apply here?

Answer 15

if you substitute x=infnity, you;'d get (infinity/infinity)
hence, you have to use L'hopital's rule

Answer 16

after applying it once, what do you get?

Answer 17

umm I got lim x-> infinity e^2x - 1/x, lim x-> infinity e^ex - 1/x, lim x -> infinity d(e^2x - 1)/dx over dx/dx = e^2

Answer 18

no comprehendo. use the equation editor below the reply box so the work is legible.

Answer 19

ok

Answer 20

\[
L=\lim_{x\to\infty}\frac{2e^{2x}}{1}=2\lim_{x\to\infty}e^{2x}\\
L=2\times e^{2\times\infty}\\
\boxed{L=\infty}
\]

Answer 21

similarly, what would happen if ...
\(x\to-\infty\)

Answer 22

ok

Answer 23

ok