$\lim_{x\to0} [\ln(1+\sin^2(x))\cdot\cot(\ln^2(1+x))]$

Question

Consider the following limit. $$\lim_{x\to0} [\ln(1+\sin^2(x))\cdot\cot(\ln^2(1+x))]$$

According to me the answer should be $0$ by directly putting the value of $x$. But the answer given as $1$. How ?

Answer 1

I thought it might be instructive to present a way forward that relies only on elementary inequalities and the squeeze theorem. To that end, we now begin with a primer.

PRIMER:

In THIS ANSWER, I developed the inequalities introduced in elementary geometry

$$\theta\cos(\theta)\le \sin(\theta)\le \theta \tag 1$$

for $0\le \theta\le \pi/2$.

And in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 2$$

for $x>0$.

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Applying $(1)$ and $(2)$ reveals

$$\frac{x^2\cos^2(x)}{1+\sin^2(x)}\le \log(1+\sin^2(x))\le \sin^2(x)\le x^2 \tag 3$$

and

$$\frac{\cos(\log^2(1+x))}{x^2}\le \cot(\log^2(1+x))\le \frac1{\log^2(1+x)}\le \frac{(1+x)^2}{x^2} \tag 4$$

Finally, using $(3)$ and $(4)$ together yields

$$ \frac{\cos^2(x)\,\cos(\log^2(1+x))}{1+\sin^2(x)}\le \log(1+\sin^2(x))\,\cot(\log^2(1+x))\le (1+x)^2$$

whence application of the squeeze theorem gives the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\log(1+\sin^2(x))\,\cot(\log^2(1+x))=1}$$

as was to be shown!

Answer 2

HINT:

$$=\lim_{x\to0}\dfrac{\ln(1+\sin^2x)}{\sin^2x}\cdot\cos(\ln^2(1+x))\cdot\left(\dfrac{\sin x}x\right)^2\cdot\dfrac1{\dfrac{\sin(\ln^2(1+x))}{{\ln^2(1+x)}}}\dfrac1{\left(\dfrac{\ln(1+x)}x\right)^2}$$

Answer 3

Observe that $\ln(1+x) \approx_{0} x, \sin x \approx_{0} x \implies \ln(1+\sin^2x) \approx_{0} x^2, \cot(\ln^2(1+x)) \approx_{0} \cot(x^2)$, and using the fact that $\dfrac{x^2}{\sin^2 x} \to 1$ as $x \to 0$, we have the limit is $1$ indeed.