1. ## limit of ln

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

2. ## Re: limit of ln

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

3. ## Re: limit of ln

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

4. ## Re: limit of ln

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

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

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

5. ## Re: limit of ln

Sorry folks, /a was dropped during Latex editing!

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# tính lim(lnx-lna)/(x-a)

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