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$$\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$$

I don't think L'hospital's rule will make the problem easy. (I am afraid to differentiate the numerator). The given limit has a $\frac{0}{0}$ form. I tried using taylor series but the it made the problem more complicated.

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    $\begingroup$ I'm afraid to differentiate the numerator as well ... Where did this monster come from? Surely not a textbook ... $\endgroup$
    – Christopher Carl Heckman
    Commented Mar 20, 2016 at 0:33
  • $\begingroup$ Is $\ln^3 (1+x)$ supposed to be $(\ln (1+x))^3$ or $\ln \ln \ln (1+x)$? If it's the latter, then the limit does not exist. If it's the former, Maple gives $2\sqrt2$. $\endgroup$
    – Christopher Carl Heckman
    Commented Mar 20, 2016 at 0:38
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    $\begingroup$ One makes the usual estimates, mostly with series. $\endgroup$
    – André Nicolas
    Commented Mar 20, 2016 at 0:39
  • $\begingroup$ @AndréNicolas : I'd LOVE to see your work this out with estimates. $\endgroup$
    – Christopher Carl Heckman
    Commented Mar 20, 2016 at 0:40
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    $\begingroup$ @CarlHeckman: The main difficulty is the typing. The $2\sqrt{2}$ comes from rationalizing the numerator in $\sin(\sqrt{x^2+2}-\sqrt{2})$. $\endgroup$
    – André Nicolas
    Commented Mar 20, 2016 at 0:43

5 Answers 5

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Use equivalents:

$$\ln(1+u)\sim_0 u, \enspace\sqrt{1+u}-1\sim_0 \dfrac u2, \enspace \sin u\sim_0 u, \enspace \cos u\sim_0 1, \enspace\cot u\sim_0 \dfrac1u, \enspace \tan u\sim_0 u$$ hence $$\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}\sim_0 \frac{x^3\cfrac1{x^3}x^4}{\sqrt2\Bigl(\sqrt{1+\frac{x^2}2}-1\Bigr)x^2}\sim_0 \frac{x^2}{\sqrt2\dfrac{x^2}{4}}=2\sqrt2.$$

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  • $\begingroup$ The second one is not an equivalent! Equivalents will by definition only "care" about the first term. I.e., writing $\sqrt{1+u} \sim_{u\to 0} 1+\frac{u}{2}$ is true, but so is $\sqrt{1+u} \sim_{u\to 0} 1+\frac{u}{\pi}$, $\sqrt{1+u} \sim_{u\to 0} 1+\sqrt{u}$, or $\sqrt{1+u} \sim_{u\to 0} 1+\frac{1}{\sqrt{\ln\ln\ln \lvert u\rvert}}$. $\endgroup$
    – Clement C.
    Commented Mar 20, 2016 at 0:58
  • $\begingroup$ I'm sorry, not particularly. We usually take as equivalents the first non-zero term in, say, Taylor's expansion, but it is in no way part of the definition of equivalents. I was too fast in adding $-1$, that is true. I've corrected it. $\endgroup$
    – Bernard
    Commented Mar 20, 2016 at 1:01
  • $\begingroup$ what is your definition of equivalent? The one I use is $f\sim_a g$ iff $f(x) - g(x) = \epsilon(x)g(x)$ for some function $\epsilon$ with $\lim_{x\to a}\epsilon(x) = 0$. (A simpler and equivalent (when the functions do not cancel on a neighborhood of $a$) is $\frac{f(x)}{g(x)} \xrightarrow[x\to a]{}1$.) $\endgroup$
    – Clement C.
    Commented Mar 20, 2016 at 1:04
  • $\begingroup$ Oh, saw your edit. Seems fine to me now. $\endgroup$
    – Clement C.
    Commented Mar 20, 2016 at 1:05
  • $\begingroup$ Exactly the same definition as you. I find more expressive to write it as $f(x)=g(x)(1+\varepsilon(x))$. $\endgroup$
    – Bernard
    Commented Mar 20, 2016 at 1:10
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Hint: as $u\to 0,$ $(\ln (1+u))/u \to 1,$ $u(\cot u) \to 1,$ $(\tan u)/u \to 1,$ $(\sin u)/u \to 1.$ No power series necessary to justify these, just well known limits and derivatives.

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This is exactly the kind of question which "looks very difficult" and at the same time is "extremely easy to answer". It appears that it is specially crafted to generate a complicated looking expression in order to intimidate a casual reader/student. The following evaluation shows that the difficulty is only superficial: \begin{align} L &= \lim_{x \to 0}\frac{\log(1 + \sin^{3}x \cos^{2}x)\cot(\log^{3}(1 + x))\tan^{4}x}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\log(1 + x^{2})}\tag{1}\\ &= \lim_{x \to 0}\dfrac{\log(1 + \sin^{3}x \cos^{2}x)\cot(\log^{3}(1 + x))\cdot\dfrac{\tan^{4}x}{x^{4}}\cdot x^{4}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\cdot\dfrac{\log(1 + x^{2})}{x^{2}}\cdot x^{2}}\notag\\ &= \lim_{x \to 0}\dfrac{\log(1 + \sin^{3}x \cos^{2}x)\cot(\log^{3}(1 + x))\cdot 1 \cdot x^{4}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\cdot 1 \cdot x^{2}}\notag\\ &= \lim_{x \to 0}\dfrac{\log(1 + \sin^{3}x \cos^{2}x)x^{2}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\tan(\log^{3}(1 + x))}\notag\\ &= \lim_{x \to 0}\dfrac{\dfrac{\log(1 + \sin^{3}x \cos^{2}x)}{\sin^{3}x \cos^{2}x}\cdot \sin^{3}x \cos^{2}x\cdot x^{2}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\cdot\dfrac{\tan(\log^{3}(1 + x))}{\log^{3}(1 + x)}\cdot \dfrac{\log^{3}(1 + x)}{x^{3}}\cdot x^{3}}\notag\\ &= \lim_{x \to 0}\dfrac{1\cdot \sin^{3}x \cos^{2}x\cdot x^{2}}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})\cdot1\cdot 1\cdot x^{3}}\notag\\ &= \lim_{x \to 0}\dfrac{\sin^{3}x \cdot 1}{\sin(\sqrt{x^{2} + 2} -\sqrt{2})x}\notag\\ &= \lim_{x \to 0}\dfrac{\dfrac{\sin^{3}x}{x^{3}} \cdot x^{3}}{\dfrac{\sin(\sqrt{x^{2} + 2} -\sqrt{2})}{\sqrt{x^{2} + 2} -\sqrt{2}}\cdot(\sqrt{x^{2} + 2} -\sqrt{2})x}\notag\\ &= \lim_{x \to 0}\dfrac{1 \cdot x^{3}}{1\cdot(\sqrt{x^{2} + 2} -\sqrt{2})x}\notag\\ &= \lim_{x \to 0}\frac{x^{2}}{\sqrt{x^{2} + 2} - \sqrt{2}}\tag{2}\\ &= \lim_{x \to 0}\frac{x^{2}(\sqrt{x^{2} + 2} + \sqrt{2})}{(x^{2} + 2) - 2}\notag\\ &= \lim_{x \to 0}(\sqrt{x^{2} + 2} + \sqrt{2})\notag\\ &= \sqrt{2} + \sqrt{2}\notag\\ &= 2\sqrt{2}\notag \end{align} The simplification of the complicated expression in equation $(1)$ to a very simple expression in equation $(2)$ is done via the use of standard limits $$\lim_{x \to 0}\cos x = \lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{\tan x}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$ In evaluation of limits it is important to figure out the use of standard limits to simplify a complicated expression.

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Using infinitesimals of equal order, $$\begin{align*} \lim_{x\to 0}\frac{\log(1+\sin^3x \cos^2x)\cot(\log(1+x)^3)\tan^4 x}{\sin(\sqrt{x^2+2}-\sqrt{2})\log(1+x^2)} &= \lim_{x\to 0}\frac{\log(1+\sin^3x \cos^2x)\cot(\log(1+x)^3)x^4}{\sin(\sqrt{x^2+2}-\sqrt{2})\log(1+x^2)} \\ &= \lim_{x\to 0}\frac{ \sin^3x \cos^2x \cot(\log(1+x)^3)x^4}{(\sqrt{x^2+2}-\sqrt{2})x^2} \\ &= \lim_{x\to 0}\frac{ x^3 \cot(\log(1+x)^3)x^4}{(\sqrt{x^2+2}-\sqrt{2})x^2} \\ &= \lim_{x\to 0}\frac{x^3 x^{-3} x^4}{(\sqrt{x^2+2}-\sqrt{2})x^2} \end{align*}$$

where $\cot(\log(1+x)^3) \sim x^{-3}$ since $\cot \sim x^{-1}, \log(1+x)^3 \sim x^3$. Some of the other infinitesimals I've used:

$$\sin x \sim x, \cos x \sim 1, \log(1+x) \sim x, \tan x \sim x $$ It is easily checked with l'Hopital that if $f \sim g$, $$ \lim_{x \to 0} \frac{f(x)}{g(x)} = 1 $$

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  • $\begingroup$ Actually, $\tan ^4 x = (\tan x)^4 = x^4 + O(x^6)$ ... [which was later fixed] $\endgroup$
    – Christopher Carl Heckman
    Commented Mar 20, 2016 at 0:39
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$\sin(1+x)=x+\mathcal O(x^3)\;$, and likewise for the logarithmic function, so:

$$\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}\cong$$

$$\frac{\left(\sin^3x\cos^2x\right)\cot\left(x^3\right)\left(x^4\right)}{\left(\sqrt{x^2+2}-\sqrt2\right)\left(x^2\right)}=\cos x^3\;\frac{x^3}{\sin x^3}\;\cos^2x\;\frac{\sin^3x}{x^3}\left(\sqrt{x^2+2}+\sqrt2\right)\longrightarrow$$

$$\xrightarrow[x\to0]{}1\cdot1\cdot1\cdot1\cdot2\sqrt2=2\sqrt2=\sqrt8$$

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