Question

Prove that lim x->0 log(1+x)/x = 1. It's urgent, please help.

Answer 1

Prove that lim x->0 log(1+x)/x.

We know log(1+x) = x - x²/2 + x³/3 - x⁴/4........

So here is your solution.

Answer 2

Lets say that we have some variable called "y" and lets set it equal to (1+x)^(1/x).

so far we have:

y=(1+x)^(1/x)

now lets take the natural log of each side to obtain

Ln(y)=Ln((1+x)^(1/x))

now, we can pull out the "1/x" term in front of the "Ln(" [because of your basic log rules] to get:

Ln(y) = (1/x)*Ln(1+x) ~ or ~ Ln(y)=Ln(1+x)/x

now we get to take the limit (yay). this leaves

Ln(y) = lim Ln(1+x)/x

~~~~~ x->0

now we get to use La Hôpital's Rule on the right hand side of the equation.

= lim (d/dx[Ln(1+x)]) / (d/dx[x])

x->0

(where d/dx[f(x)] refers the the derivative of f(x) with respect to "x")

= lim (d/dx[1+x]/(1+x)) / 1

x->0

= lim 1/(1+x)

x->0

This last Limit can be simply evaluated by just plugging in 0 for "x"

= lim 1/(1+0) = 1

x->0

So! remembering the left hand side of the equation, we have

Ln(y) = lim Ln(1+x)/x

~~~~~x->0

Ln(y)=1 ~and~ y = e^1 = e

Thus: Lim (x+1)^(1/x)=e

Q.E.D