limit of ln

**calman** · October 7th 2012, 06:10 PM

What is the limit as x approaches a for, (ln(x)-ln(a))/(x-a)

**MaxJasper** · October 7th 2012, 08:07 PM

$\lim_{x\to a} \, \frac{\log (x)-\log (a)}{x-a}=1$

**hollywood** · October 8th 2012, 06:58 AM

Isn't that the definition of the derivative of $\log{x}$ at $x=a$ ? So it would be 1/a.

**johnsomeone** · October 8th 2012, 11:48 AM

Originally Posted by calman

What is the limit as x approaches a for, (ln(x)-ln(a))/(x-a)

As Hollywood posted, they want you to see that it's the derivative of f(x) = ln(x) at x = a, and so is 1/a. I've usually seen that done by defining the natural log as an integral, and then taking its derivative using the Fundamental Theorem of Calculus.

However, if you only know about the derivative of $e^x$ , and have $ln(x)$ defined as its inverse, you could do this:

Let $f(x) = e^x$ . Since $f^{-1}(f(x)) = x \ \forall x \in \mathbb{R}$ , take derivatives to get $(f^{-1})'(f(x))(f'(x)) = 1 \ \forall x \in \mathbb{R}$ .

Since $f(x) = e^x$ , that reads $(f^{-1})'(f(x)) = 1/f(x) \ \forall x \in \mathbb{R}$ , so

$(f^{-1})'(e^x) = 1/(e^x) \ \forall x \in \mathbb{R}$ , so, since the image of f is all positive reals, have

$(f^{-1})'(t) = 1/t \ \forall t> 0, t \in \mathbb{R}$ .

Also, could do this:

$\lim_{x \to a} \frac{ln(x)-ln(a)}{x-a} = \lim_{x \to a} \frac{ln(x/a)}{a\left(\frac{x}{a}-1\right)}$

$= \frac{1}{a} \ \lim_{x \to a} \frac{ln(x/a)}{\frac{x}{a}-1} = \frac{1}{a} \ \lim_{t \to 1} \frac{ln(t)}{t-1}$

$=$ (substituting $t = e^s$ ) $\frac{1}{a} \ \lim_{s \to 0} \frac{ln(e^s)}{e^s-1} = \frac{1}{a} \ \lim_{s \to 0} \frac{s}{e^s-1}$

$= \frac{1}{a} \ \lim_{s \to 0} \frac{1}{\frac{e^s-1}{s}} = \frac{1}{a} \ \frac{1}{\lim_{s \to 0} \frac{e^s-1}{s}}$

$= \frac{1}{a} \ \frac{1}{1} = \frac{1}{a}$

**MaxJasper** · October 8th 2012, 12:11 PM

Sorry folks, /a was dropped during Latex editing!

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