Question

limit as x approaches 0. (e^(2x) -1) / ( 1- e^X)

Answer 1

see if u can factor numerator

Answer 2

e^(2x) -1 = (e^x)^2 - 1^2

Answer 3

ok

Answer 4

what should i do next ??

Answer 5

do u knw this identity ?

\(a^2 - b^2 = (a+b)(a-b)\)

Answer 6

yes

Answer 7

\(\large \lim \limits_{x\to 0} ~\frac{e^{2x}-1}{1-e^x}\)

\(\large \lim \limits_{x\to 0} ~\frac{(e^{x})^2-1^2}{1-e^x}\)

Answer 8

apply that identity to numerator

Answer 9

\(\large \lim \limits_{x\to 0} ~\frac{e^{2x}-1}{1-e^x}\)

\(\large \lim \limits_{x\to 0} ~\frac{(e^{x})^2-1^2}{1-e^x}\)

\(\large \lim \limits_{x\to 0} ~\frac{(e^x+1)(e^x-1)}{1-e^x}\)

Answer 10

can u do the rest ha ?

Answer 11

so now we can eliminate 1-e^X right ?

Answer 12

yes thats the reason we factored :)

Answer 13

do it, and let me knw the final answer u get

Answer 14

is it 2

Answer 15

close, but no. try again

Answer 16

\(\large \lim \limits_{x\to 0} ~\frac{e^{2x}-1}{1-e^x}\)

\(\large \lim \limits_{x\to 0} ~\frac{(e^{x})^2-1^2}{1-e^x}\)

\(\large \lim \limits_{x\to 0} ~\frac{(e^x+1)(e^x-1)}{1-e^x}\)

\(\large \color{red}{-}\lim \limits_{x\to 0} ~\frac{(e^x+1)(e^x-1)}{e^x-1}\)

\(\large \color{red}{-}\lim \limits_{x\to 0} ~e^x+1\)

take the limit now

\(\large \color{red}{-} (e^0 + 1)\)

\(\large \color{red}{-} (1 + 1)\)

\(\large \color{red}{-} (2)\)

\(\large -2\)

Answer 17

thanks a lot :)

Answer 18

np.. u wlc :)