# Question 46f42

Feb 12, 2017

${\lim}_{x \to 0} \frac{\ln \left(1 + 3 x\right)}{\sin} \left(x\right) = 3$

#### Explanation:

${\lim}_{x \to 0} \frac{\ln \left(1 + 3 x\right)}{\sin} \left(x\right)$

For starters, we can see that if we set $x = 0$ we are looking at:

$\ln \frac{1}{\sin} \left(0\right) = \frac{0}{0}$.

That is in indeterminate form and so L'Hôpital's Rule can be used. That can be a bit of a mechanical black-box and results, after one application, in this:

$= {\lim}_{x \to 0} \frac{\frac{3}{1 + 3 x}}{\cos} \left(x\right) = 3$

If you are familiar with Taylor Series, we can look at this in a bit more detail:

\sin (x) \approx x- x^{3}/{3!}+x^{5}/{5!}- x^{7}/{7!}

....and...

$\log \left(1 + x\right) = x - {x}^{2} / 2 + {x}^{3} / \left\{3\right\} + O \left({x}^{4}\right)$

Or:

$\log \left(1 + 3 x\right) = 3 x - {\left(3 x\right)}^{2} / 2 + {\left(3 x\right)}^{3} / \left\{3\right\} + O \left({x}^{4}\right)$

As $x \to 0$, we can ignore ${x}^{2}$ higher order terms and compare the first terms (in ${x}^{1}$) to conclude the same thing, namely that:

${\lim}_{x \to 0} \frac{\ln \left(1 + 3 x\right)}{\sin} \left(x\right) = {\lim}_{x \to 0} \frac{3 x}{x} = 3$

It might then be helpful to show how the Taylor Series connect into L'Hôpital, if we compare series for 2 functions, f(x) and g(x), about points at which $f \left(a\right) = g \left(a\right) = 0$.

They create the following ratio:

(f(x))/(g(x)) = (f(a)+ {f'(a)}/{1!}(x-a)+ {f''(a)}/{2!}(x-a)^{2}+\cdots )/(g(a)+ {g'(a)}/{1!}(x-a)+ {g''(a)}/{2!}(x-a)^{2}+\cdots)#

For $f \left(a\right) = g \left(a\right) = 0$ and $x \approx a$ you can see already that the predominant terms in this ratio are the first derivatives.