Evaluate the limit $\lim_{x\to 0} \frac{\ln |1+x^3|}{\sin^3 x}$

Question

I tried this with L’Hopital and got $1$ as the answer. I don’t know if it can be applied considering the presence of the modulus function, but the answer is right. But I want a solution without using that rule, and I don’t know how to start this. Can I get some insight into this?

Edit: MY ATTEMPT

$$\lim_{x\to 0} \frac{\ln |1+x^3|}{|1+x^3|}\lim_{x\to 0} \frac{|1+x^3|}{\sin^3x}$$ $$=\lim _{x\to 0} \frac{x^3|\frac{1}{x^3}+1|}{\sin^3x}$$

Answer 1

It comes down to two known limits as follows: $$\lim_{x\to 0} \frac{\ln |1+x^3|}{\sin^3 x}=\lim_{x\to 0} \frac{\ln |1+x^3|}{x^3}\frac{x^3}{\sin^3 x}=1\cdot 1=1$$

Answer 2

1. if $|x|$ is "small", then $1+x^3 >0.$ Hence we have to compute

$$\lim_{x\to 0} \frac{\ln (1+x^3)}{\sin^3 x}.$$

2. $\frac{\ln (1+x^3)}{\sin^3 x}= \frac{x^3}{\sin^3 x} \cdot \frac{\ln (1+x^3)}{x^3}.$

Can you proceed ?

Answer 3

$$\ln(1+z)=z-z^2/2+\cdots, \sin z=z-z^3/6+\cdots$$ $$\lim_{x\to 0} \frac{x^3-x^6/2+\cdots}{(x-x^3/6+\cdots)^3} =\lim_{x \to 0} \frac{1-x^3/2+\cdots}{1-3x^2/6+\cdots}=1.$$ Lastly, we have used the binomial series: $(1+z)^{\nu}=1+\nu z+\cdots$, if $|z|<1$.

Answer 4

First observe the absolute value isn't necessary, as $1+x^3>0$ if $-1<x<1$. You can find the limit very with equivalents near $0$: we know that $\sin\sim_0 x$ and $\ln(1+u)\sim_0 u $, whence $$\frac{\ln(1+x^3)}{\sin^3x}\sim_0 \frac{x^3}{x^3}=1.$$