# Thread: Limits - Explanation of a step.

1. ## Limits - Explanation of a step.

Perhaps someone could explain the last step here for me please...

Find the limit of $\mathop {\lim}\limits_{x \to 1} \frac{In(x)}{x-1}$. Do not use L'Hopital's rule.

Solution;
Since In(1) = 0, we have that,

$\mathop {\lim}\limits_{x \to 1} \frac{In(x)}{x-1}$ = $\mathop {\lim}\limits_{x \to 1} \frac{In(x) - In(1)}{x-1}$ = $\frac{d}{dx} In(x) = 1$ for x =1. (Not quite sure how to do the latex for the way its written in my notes. To the right of the d/dx is a vertical line with x=1 at the bottom.)

And... Im not quite sure how to do the last step there i.e the $\mathop {\lim}\limits_{x \to 1} \frac{In(x) - In(1)}{x-1}$ = $\frac{d}{dx} In(x) = 1$ bit...

Any help on this?

2. First off, the natural logarithm is abbreviated $\ln (x)$ (an L not an I).

Recall the definition of a derivative of a function at some point x = a:
$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

So imagine: $f(x) = \ln (x)$ and a = 1.

3. jeez thats so simple..! Cheers for that.

a) Let $u=x-1,$ and the limit becomes $\underset{u\to 0}{\mathop{\lim }}\,\frac{\ln (1+u)}{u}=\underset{u\to 0}{\mathop{\lim }}\,\ln (1+u)^{1/u}.$
b) Let $z=\frac1u,$ and the limit becomes $\underset{z\to \infty }{\mathop{\lim }}\,\ln \left( 1+\frac{1}{z} \right)^{z}=\ln e=1.$