Question

Lim (e^(2x) - 1)/x when x goes to 0 whithout L'hospital...

Answer 1

it will be the ratio of differentiation of e^(2x)-1 and x which is = 2*e^(2x)/1 now when x tends to zero it will be equal to 2

Answer 2

But, this is l'hospital right? What I need, please, is using pure algebra. I think that I need to use the fundamental limit that equals to e.

Answer 3

yes here i've use l'hospital's rule

Answer 4

well i don't understand the use of pure algebra here...and if you want to solve this question you can do it by the fundamental definition of limit

Answer 5

\[\lim_{x\to0}{e^{2x}-1\over x}=\exp\left(\lim_{x\to0}\ln\left({e^{2x}-1\over x}\right)\right)\]\[=\exp\left(\lim_{x\to0}[\ln(e^x-1)+\ln(e^x+1)-\ln x]\right)\]then what else perhaps I have no idea...

Answer 6

Put the limits:

\[=\exp\left([\cancel{\ln(0)}+\ln(1+1)-\cancel{\ln (0)}]\right) \implies \exp(\ln(2)) = 2\]

Ha ha ha..

Answer 7

I know, if only :)

Answer 8

@waterineyes how did you calculate ln(0) i guess it is not defined and what you have done here is you have subtracted ln(0) from the value of ln(1) which is zero

Answer 9

How ln1 ??

Answer 10

in ln(e^x-1) if you'll put x=0 you'll get e^0=1 and 1-1=0 so ln(0) is there now how you have eliminated ln(0)

Answer 11

\[\ln(e^x - 1) \implies \ln(e^0 - 1) \implies \ln(1 - 1) = \ln(0)\]

Answer 12

Last term is also ln(0) so if they are same then you can cancel them..

Answer 13

yea sorry my bad i got confused with the second term