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

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

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.

• I'm afraid to differentiate the numerator as well ... Where did this monster come from? Surely not a textbook ... Commented Mar 20, 2016 at 0:33
• 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 $2\sqrt2$. Commented Mar 20, 2016 at 0:38
• One makes the usual estimates, mostly with series. Commented Mar 20, 2016 at 0:39
• @AndréNicolas : I'd LOVE to see your work this out with estimates. Commented Mar 20, 2016 at 0:40
• @CarlHeckman: The main difficulty is the typing. The $2\sqrt{2}$ comes from rationalizing the numerator in $\sin(\sqrt{x^2+2}-\sqrt{2})$. Commented Mar 20, 2016 at 0:43

## 5 Answers

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

• 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 $\sqrt{1+u} \sim_{u\to 0} 1+\frac{1}{\sqrt{\ln\ln\ln \lvert u\rvert}}$. Commented Mar 20, 2016 at 0:58
• 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. Commented Mar 20, 2016 at 1:01
• 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 $\frac{f(x)}{g(x)} \xrightarrow[x\to a]{}1$.) Commented Mar 20, 2016 at 1:04
• Oh, saw your edit. Seems fine to me now. Commented Mar 20, 2016 at 1:05
• Exactly the same definition as you. I find more expressive to write it as $f(x)=g(x)(1+\varepsilon(x))$. Commented Mar 20, 2016 at 1:10

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.

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.

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 $f \sim g$, $$\lim_{x \to 0} \frac{f(x)}{g(x)} = 1$$

• Actually, $\tan ^4 x = (\tan x)^4 = x^4 + O(x^6)$ ... [which was later fixed] Commented Mar 20, 2016 at 0:39

$\sin(1+x)=x+\mathcal O(x^3)\;$, and likewise for the logarithmic function, so:

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

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

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