$\ln$ and $\ln^2$

We could solve this problem using a variety of methods, from Expansions to L'Hospital's Rule.

I will do it the easiest way - using expansions.

As $x \to 0$ , this is what happens

• $\ln (1 + \sin x) \approx \sin x$

• $\sin x \approx x$

$2 - \cos2x = 1 + 2\sin^2x$

Using the above information,

$\ln(2- \cos2x) \approx \ln(1+2\sin^2x) \approx 2\sin^2x \approx 2x^2$

$\ln^2(\sin3x + 1) = \ln(\sin3x + 1) \ln(\sin3x + 1) \approx \sin3x \sin3x \approx 9x^2$

Hence the required limit reduces to:

$\frac{a}{b} = \frac{2x^2}{9x^2} = \frac{2}{9}$

$a^2 + b^2 = \boxed{85}$

I hope I got the parentheses correct!

If you try substituting zero initially, you get 0/0. So, L'Hopital's rule can be applied here. $\displaystyle = \lim_{x \rightarrow 0} \displaystyle \frac{\frac{1}{2-\cos(2x)}\cdot \sin(2x) \cdot 2}{2\ln(\sin(3x) + 1) \cdot \frac{1}{\sin(3x)+1}\cdot \cos(3x) \cdot 3}$ If you try substituting again, you still get 0/0 so you must apply L'Hopital's rule again. $\displaystyle = \lim_{x \rightarrow 0}\frac{\displaystyle\frac{2(\cos(2x)(2-\cos(2x)-(\sin(2x))(\sin(2x))(2))}{(2-\cos(2x))^{2}}}{\displaystyle \frac{((\sin(3x)+1)(\cos(3x)\cdot \frac{1}{\sin(3x)+1}\cdot \cos(3x) \cdot 3)+(3 \cdot -\sin(3x))(\ln(\sin(3x+1)))-(\cos(3x))(\ln(\sin(3x)+1))(\cos(3x)))\cdot3}{(\sin(3x)+1)^{2}}}$

Substituting in $0$ , we know that $\sin(a \cdot 0) = 0$ and $\cos (a \cdot 0) = 1$ so you end up with $\frac{\frac{2}{1}}{\frac{9}{1}}=\frac{2}{9}$ and $2^{2}+9^{2} = \boxed{85}$