Calculate limit: $\lim_{x\to0} \frac{\log(1+\frac{x^4}{3})}{\sin^6(x)}$ without de l'Hôpital rule

Question

I want to calculate the limit:

$$\lim_{x\to0} \frac{\log(1+\frac{x^4}{3})}{\sin^6(x)} $$

Obviously the Indeterminate Form is $\frac{0}{0}$.

I've tried to calculate it writing:

$$\sin^6(x)$$

as

$$\sin^2(x) \cdot \sin^2(x) \cdot \sin^2(x)$$

then applying de l'Hôpital rule to the limit:

$$\lim_{x\to0} \frac{\log(1+\frac{x^4}{3})}{\sin^2(x)} \stackrel{\text{de l'Hôpital}}= \lim_{x\to0} \frac{\frac{4x^3}{x^4 + 3}}{2 \cdot \sin x \cdot \cos x} \stackrel{\text{de l'Hôpital}}= \frac{\frac{12x^2(x^4+3)-16x^6}{(x^4+3)^2}}{2(\cos^2x-\sin^2x)} \stackrel{\text{substituting x}}= \frac{0}{2(\cos^20-\sin^20)} = \frac02 = 0$$

but multiplying with the remaining two immediate limits, using the fact that the limit of a product is the product of the limits, I still have indeterminate form:

$$\lim_{x\to0} \frac{\log(1+\frac{x^4}{3})}{\sin^6(x)} = \lim_{x\to0} \frac{\log(1+\frac{x^4}{3})}{\sin^2(x)} \cdot \lim_{x\to0} \frac{1}{\sin^2x} \cdot \lim_{x\to0} \frac{1}{\sin^2x} = 0 \cdot \infty \cdot \infty = \infty $$

I was wondering if there was a (simpler?) way to calculate it without using de l'Hôpital rule...

... Maybe using some well known limit? But I wasn't able to figure out which one to use and how to reconduct this limit to the well know one...

Can you please give me some help? Ty in advice as always

Answer 1

By first order expansion $$\frac{log(1+\frac{x^4}{3})}{sen^6(x)} =\frac{\frac{x^4}{3}+o(x^6)}{x^6+o(x^6)} \to +\infty$$

By standard limits $$\frac{log(1+\frac{x^4}{3})}{\frac{x^4}{3}} \frac{\frac{x^4}{3}}{x^6}\frac{x^6}{sen^6(x)}\to +\infty$$

Answer 2

Hint:

Use the basic formulae: $$\lim_{h\to0}\dfrac{\ln(1+h)}h=1$$ $$\lim_{u\to0}\dfrac{\sin u}u=1$$

Answer 3

$$\lim_{x\rightarrow0}\frac{\ln\left(1+\frac{x^4}{3}\right)}{\sin^6x}=\lim_{x\rightarrow0}\left(\left(\frac{x}{\sin{x}}\right)^6\cdot\ln\left(1+\frac{x^4}{3}\right)^{\frac{3}{x^4}}\cdot\frac{1}{3x^2}\right)=+\infty$$