Solve limit $\lim_{x \rightarrow 0}\frac{x^{2}\ln\left(1+2x\right)}{2\sin\left(x\right)\left[ \cos\left(3x\right)-1\right]} $ without L'Hopital

Question

I am trying to find out a way to calculate this limit. I tried L'Hopital but still give me indetermined form.

$\lim_{x \rightarrow 0}\frac{x^{2}\ln\left(1+2x\right)}{2\sin\left(x\right)\left[ \cos\left(3x\right)-1\right]} $

Answer 1

Apply Hospitals a 2nd and a 3rd time.

Or do a Taylor series expansion.

$\frac {-2x^3 + o(x^4)}{2 (x + o(x^3))(-\frac 12 (3x)^2 + o(x^4))}$

$-\frac 29$

Answer 2

Equivalents will do:

• $\ln(1+u)\sim_0 u$,
• $\cos u-1\sim_0-\dfrac{u^2}2$,
• $\sin u\sim_0 u$, hence $$\frac{x^{2}\ln\left(1+2x\right)}{2\sin\left(x\right)\left[ \cos\left(3x\right)-1\right]}\sim_0\frac{x^{2}\cdot 2x}{2x\Bigl( -\dfrac{9x^2}2\Bigr)}=-\frac 29.$$

Answer 3

Write as $$\frac{1}{2}\frac{x}{\sin x}\frac{1}{9}\frac{(3x)^2}{\cos 3x-1}2\frac{\ln (1+2x)}{2x}$$ Limit $-\frac{2}{9}$.

Answer 4

Since $$\lim_{x\to0}\frac{sin(x)}{x}=1$$ $$\lim_{x\to0}\frac{ln(1+x)}{x}=1$$ $$\lim_{x\to0}\frac{1-cosx}{x^2}=\frac{1}{2}$$ the above can be simplified as: $$\lim_{x\to0}\frac{\frac{ln(1+2x)}{2x}\times2}{2\times\frac{\sin(x)}{x}\times\frac{cos(3x)-1}{(3x)^2}\times3^2}$$ $$=\frac{1\times2}{2\times1\times-\frac{1}{2}\times9}=-\frac{2}{9}$$