Can anyone solve this problem?

Question

Evaluate: lim (e^3x - 1)/(e^x - 1)
.............. .. x=>0

Answer 1

Ok, here is the right answer...

first, look at what you are looking to find limes for:

a(x) = (e^3x-1)/(e^x-1) = [ (e^x)^3 - 1 ] / (e^x - 1) =

= [ (e^x-1)( (e^x)^2 + e^x + 1)]/(e^x-1) = (e^x)^2 + e^x + 1 =

= e^(2x) + e^x + 1.

therefore,
lim(a(x))=e^0 + e^0 + 1 = 3, when x-> 0

Answer 2

First let meexplain for you something;
if there are two functions and Lim f(x) =L when x –> x0 and Lim g(x) = M when x –> x0 ( L & M are real numbers ) so we have ;
1) Lim f(x)g(x) = L * M
2) Lim f(x) / g(x) = L/M , M ≠0

so lets start,
if f(x) = (e^3x - 1) & g(x) =(e^x - 1)
lim (e^3x - 1)/(e^x - 1) =[ lim (e^3x - 1) ] / [ lim (e^x - 1)]
x –> 0
[ (lim e^3x ) * (lim e^-1) ] / [ (lim e^x)* (lim e^-1 )]=
[ (lim e^3*0 ) * (1/e) ] / [ (lim e^0)* (1/e)]=
(1+ 1/e) /(1+ 1/e) = 1

lim (e^3x - 1)/(e^x - 1) = 1
x –> 0
Good Luck.

Answer 3

e^3x-1=o
e^3x=1

Answer 4

limit doesnt exist, they approch different numbers from each side