Evaluating limit with natural logarithm: h-->0 of [ln(2+h) - ln2]/h

[lhagen]

Use definition of the derivative to evaluate the lim h-->0 of [ln(2+h) - ln2]/h
I have tried breaking up the function but I cannot get past finding indeterminate solutions.

 

[pka]

Use definition of the derivative to evaluate the lim h-->0 of [ln(2+h) - ln2]/h
I have tried breaking up the function but I cannot get past finding indeterminate solutions.

The idea is simple:
$\displaystyle \displaystyle{\lim _{h \to 0}}\frac{{\log (2 + h) - \log (2)}}{h} = {D_{x = 2}}\left[ {\log (x)} \right]$

 

[lookagain]

1. ------------------------------------------------
   
   "Use definition of the derivative" is what the problem wants as the method.
   
   ​

$\displaystyle \displaystyle\lim_{h\to 0} \ \dfrac{1}{h}ln\bigg(\dfrac{2 + h}{2}\bigg) \ =$


$\displaystyle \displaystyle\lim_{h\to 0} \ ln\bigg[\bigg(1 + \dfrac{h}{2}\bigg)^{1/h}\bigg]$



Let h/2 = x


Then h = 2x


And 1/h = 1/(2x)


As x approaches 0, 2x approaches 0, and x approaches 0.


2. Substitute these:


$\displaystyle \displaystyle\lim_{x\to 0} \ ln\bigg[(1 + x)^{\tfrac{1}{2x}}\bigg] \ =$

$\displaystyle \displaystyle\lim_{x\to 0} \ ln\bigg[(1 + x)^{\tfrac{1}{x}}\bigg]^{\tfrac{1}{2}} \ =$

$\displaystyle \displaystyle\lim_{x\to 0} \ \dfrac{1}{2}ln\bigg[(1 + x)^{\tfrac{1}{x}}\bigg] \ =$

$\displaystyle \dfrac{1}{2}\displaystyle\lim_{x\to 0} \ ln\bigg[(1 + x)^{\tfrac{1}{x}}\bigg] \ =$

$\displaystyle \dfrac{1}{2}ln\bigg\{\displaystyle\lim_{x\to 0} \bigg[(1 + x)^{\tfrac{1}{x}}\bigg]\bigg\} \ =$


Can you finish?