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I'm preparing for my first exam in university (just recently enrolled in computer science) and I'm having difficulties working out this limit. I either currently lack the proper reasoning process to get it done or they haven't yet explained us all the theorems needed. I'd be really grateful if someone could point me in the right direction. Thank you!

$\lim\limits_{x \to 0} \frac{x^2 \sin(2x)}{\log (1+(\sin3x)^3)}$

Samuele B.
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    You can always apply L hopital rule in ${0 \over 0} $ and ${\infty \over \infty}$ forms. – Rishi Oct 12 '19 at 13:06
  • Just apply the asymptotics $\sin{(x)}\sim x$ and $\log{(1+x)}\sim x$ as $x\to0$ which directly gives the limit as $2/27$. – Peter Foreman Oct 12 '19 at 13:08
  • @Peter Foreman, I think I almost grasp what you said but not quite totally. Would you mind going just a tad in depth with it or even just link something that explains the concept of what you did? Thank you – Samuele B. Oct 12 '19 at 14:51
  • @SamueleB. Near $x=0$ the functions $\sin{(x)}$ and $\log{(1+x)}$ behave almost exactly like the function $x$. We can in this case just replace these two functions by $x$ in the limit to resolve it. All I'm doing here is applying the Taylor series expansion of each function with the first non-zero terms. – Peter Foreman Oct 12 '19 at 15:07
  • @Peter Foreman thank you for clarifying. I have never heard of the Taylor series. I looked it up and it uses derivatives, which we haven't covered yet in our course. Is it possible that, not having covered derivatives yet, that limit wasn't solvable with the current knowledge from the course? The exercises weren't rated in difficulty, so I have no way of knowing if I was supposed to be able to solve it yet or not for now. – Samuele B. Oct 12 '19 at 16:49

3 Answers3

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We have that

$$ \frac{x^2 \sin(2x)}{\log (1+(\sin3x)^3)}=\frac{(\sin(3x))^3}{\log (1+(\sin3x)^3)}\cdot \frac{(3x)^3 }{(\sin3x)^3}\cdot \frac{ \sin(2x)}{2x}\cdot \frac2{27}$$

then refer to standard limits as $u \to 0$

  • $\frac{\log (1+u)}{u}\to 1$
  • $\frac{\sin u}{u}\to 1$
user
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  • Care to explain how you turned the expression into what's on the right side of the equal? I can't seem to understand it. – Samuele B. Oct 12 '19 at 13:15
  • @SamueleB. If we multiply the terms on the RHSwe obtain the original expression on the LHS. The idea is to factorize the original limit by factors for which we can apply standard limits. Of course it is not always possible but in general it is the first try to do. – user Oct 12 '19 at 13:17
  • Okay now it makes sense, thank you! As a follow-up question, since I can now see what you did looking "in reverse" but I doubt I would be able to do it again from scratch, what's the order in which you did all the factorization to get from the LHS to the RHS? I hope the question is clear. I'm interested in what the thought process is, starting from when you are presented with the expression – Samuele B. Oct 12 '19 at 13:22
  • @SamueleB. You only need to practice with this method. It is the first try to do when we face with an indeterminat e form. It is not so difficult, we need to know very well the standard limits and factorize the expression. – user Oct 12 '19 at 13:27
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Hint: By L'Hôpital ("$\frac{0}{0}$"), \begin{align} \lim_{x \to 0} \frac{x^2 \sin(2x)}{\log(1 + \sin^3(3x))} & = \lim_{x \to 0} \frac{2(x^2 \cos(2x)+ x \sin(2x))(\sin^3(3x) + 1)}{9 \cos(3x) \sin^2(3x)} \\ & = \frac{2}{9} \left( \lim_{x \to 0} \left(\frac{x}{\sin(3x)}\right)^2 + \lim_{x \to 0} \left(\frac{x \sin(2x)}{\sin^2(3x)}\right)\right) \\ &= \frac{2}{9} \left( \frac{2}{9} + \frac{1}{9}\right) = \frac{2}{27}. \end{align}

ViktorStein
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Let use two results if $t \to 0$ then ${\log (1+t) \over t}=1$ and ${\sin t \over t}=1$

Now give everyone what they want

$$x^2.2x.{\sin 2x \over 2x}.\frac{\sin^3x}{\log (1+\sin^33x)}. \frac{(3x)^3}{\sin^33x}.{1 \over (3x)^3}\\={2 \over 27}$$

Assume $sin^3x$ or $2x$ as t . You will see . For exam you can remember this like sin zero upon same zero = 1 . Or $\lim_{f(x) \to 0}{\sin(f(x)) \over f(x)}=1$ and $\lim_{f(x) \to 0}{\log(1+f(x)) \over f(x)}=1$

MafPrivate
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Rishi
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