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:



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

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

  • 1
    $\begingroup$ You mean (in English) sin, not sen. $\endgroup$
    – KCd
    Dec 19 '17 at 0:37
  • $\begingroup$ Sorry, copying from italian. Edited $\endgroup$
    – ela
    Dec 19 '17 at 0:46
  • $\begingroup$ The step where you say "infinite order comparing" is incorrect. $\lim_{x\to0}\frac{\frac{12x^2(x^4+3)-16x^6}{(x^4+3)^2}}{2(\cos^2x-\sin^2x)}$ is just $0$, not $\infty$ as you ultimately find. ($\frac{\frac{12\cdot0(0+3)-16\cdot0^6}{(0+3)^2}}{2(1-0)}=0$). This further invalidates future steps, where you would be multiplying $0\cdot\infty\cdot\infty$. $\endgroup$ Dec 19 '17 at 0:51
  • $\begingroup$ @AlessioMartorana If you are ok, you can set as solved. Thanks! $\endgroup$
    – user
    Dec 19 '17 at 15:54
  • $\begingroup$ @alex.jordan You were right, fixed $\endgroup$
    – ela
    Dec 20 '17 at 0:21


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

  • $\begingroup$ That's the simplest way. $\endgroup$
    – zhw.
    Dec 20 '17 at 1:24

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




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