Logarithms Limits

$\lim_{x\to 0} \dfrac {\ln (x+1)^3}{x^2 +6x}$

$\lim_{x\to 0} \dfrac {3 \ln (x+1)}{x^2 +6x}$

Using L'Hopitals rule :

$\lim_{x\to 0} \dfrac {3 \times \dfrac {1}{(x+1)}}{2x+6}$

Now putting $x=0$ , we get $\lim_{x \rightarrow 0} {\dfrac{\ln{(x^3 + 3x^2 + 3x + 1)}}{x^2 + 6x}} = \dfrac {1}{2}= 0.5$

We can write the given fraction as

$\frac{ln((x+1)^3)}{x(x+6)}$

= $\frac{3}{x+6}$ $*$ $\frac{ln(x+1)}{x}$

Taking limit at x going to 0, the second fraction goes to 1, according to the standard result on logarithmic limits , and the first fraction goes to $\frac{1}{2}$ .

First step for $\lim_{x\to 0} \left(\frac{\ln(x^3+3x^2+3x+1)}{x^2+6x}\right)$

Steps 1.Use L'Hopital's rule

2.Find $\frac{d}{dx}(\ln(x^3+3x^2+3x+1))$ and $\frac{d}{dx}(x^2+6x)$

3. $\frac{3}{6}$ reduce into $\frac{1}{2}$

$\square$

$\LARGE \text{FIN!!!}$