(e^2x-1)/x where lim x tends to 0

Question

limits of exponential and logarithmic functions

Answer 1

(e^2x-1)/x where lim x tends to 0

The expansion for e^x is: e^x = 1 + x + x^2/2! + ... + x^n/n!

Therefore,e^(2x) is: e^(2x) = 1 + (2x) + (2x)^2/2! + ... + (2x)^n/n!

Giving  e^(2x) - 1 = (2x) + (2x)^2/2! + ... + (2x)^n/n!

Then (e^(2x) - 1)/x = (2x + 4x^2/2! + ... + 2^n.x^n/n!)/x

i.e. (e^(2x) - 1)/x = (2 + 4x/2! + ... + 2^n.x^(n-1)/n!)

So Limit, as x -> 0 = 2