Evaluation of limit at infinity: $\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$

Question

$$\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$$

What I tried was writing $1/x=t$ and making the limit tend to zero and writing the cos term in the form of sin

Answer 1

Here, we present a way forward that does not rely on differential calculus. Rather, we use only elementary inequalities and the squeeze theorem. To that end, we proceed.

Inequalities 1:

Recall from basic geometry, that the sine function is bounded as

$$\bbox[5px,border:2px solid #C0A000]{\theta \cos(\theta)\le \sin(\theta)\le \theta} \tag 1$$

for $0\le \theta\le \pi/2$ (SEE THIS ANSWER).

Letting $\theta = \arccos(t)$ in $(1)$ reveals

$$\bbox[5px,border:2px solid #C0A000]{\sqrt{1-t^2}\le \arccos(t)\le \frac{\sqrt{1-t^2}}{t}} \tag 2$$

for $0<t\le 1$.

Inequalities 2:

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

$$\bbox[5px,border:2px solid #C0A000]{\frac{t-1}{t}\le \log(t)\le t-1} \tag 3$$

for $t>0$.

Let $L$ be defined as the limit

$$L=\lim_{x\to \infty}x^2\sin\left(\log\left(\cos^{1/2}(\pi/x)\right)\right)$$

Enforce the substitution $x=\frac{1}{\pi}\arccos(t)$. Then, we can express the limit $L$ as

$$L=\lim_{t\to 1^-}\pi^2 \frac{\sin\left(\frac12\log\left(t\right)\right)}{\arccos^2(t)}$$

Applying inequalities $(1)-(3)$ we find that

$$\pi^2 \left(\frac{\frac12 \cos(\log(t))\,\frac{t-1}{t}}{\frac{1-t^2}{t^2}}\right)\le \pi^2 \frac{\sin\left(\frac12\log\left(t\right)\right)}{\arccos^2(t)}\le \pi^2 \left(\frac{\frac12 (t-1)}{(1-t^2)}\right)$$

or after simplifying

$$-\left(\frac{\pi^2}{2(1+t)}\right)t\,\cos(\log(t))\le \pi^2 \frac{\sin\left(\frac12\log\left(t\right)\right)}{\arccos^2(t)}\le -\left(\frac{\pi^2}{2(1+t)} \right) \tag 4$$

Applying the squeeze theorem to $(4)$ yields the coveted equality

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}x^2\sin\left(\log\left(\cos^{1/2}(\pi/x)\right)\right)=-\frac{\pi^2}{4}}$$

Answer 2

Using Taylor series for $\cos u$, $\sqrt{1+u}$, $\ln(1+u)$, and $\sin u$ when $u\to 0$. When $x\to \infty$, $\frac{1}{x} \to 0$, so

$$\cos\frac{\pi}{x} = 1- \frac{\pi^2}{2x^2} + o\left(\frac{1}{x^2}\right)$$ and $$\sqrt{ \cos(\frac{\pi}{x}) } = \sqrt{ 1- \frac{\pi^2}{2x^2} + o\left(\frac{1}{x^2}\right) } = 1- \frac{\pi^2}{4x^2} + o\left(\frac{1}{x^2}\right)$$ so that $$\ln \sqrt{ \cos(\frac{\pi}{x}) } = \ln\left (1- \frac{\pi^2}{4x^2} + o\left(\frac{1}{x^2}\right) \right) = - \frac{\pi^2}{4x^2} + o\left(\frac{1}{x^2}\right)$$ and finally $$ x^2 \sin \ln \sqrt{ \cos(\frac{\pi}{x}) } = x^2 \sin\!\left( - \frac{\pi^2}{4x^2} + o\!\left(\frac{1}{x^2}\right)\right) = -x^2 \left(\frac{\pi^2}{4x^2} + o\!\left(\frac{1}{x^2}\right)\right) = - \frac{\pi^2}{4} + o(1)$$

Answer 3

$\ln(\cos(\pi /x)^{1/2}) = 1/4\, \ln(\cos (\pi /x)^2) = 1/4\, \ln(1 - (\sin (\pi /x))^2)$. Also, $\lim_{z \to 0}\frac {\sin z} z = 1$ and $\lim_{u \to 0 }\frac {\ln(1-u)} u = -1$ (for example by l'Hopital). Now:

$$x^2 \sin(\ln(\cos(\pi/x)^{1/2})) = \\[12pt] x^2\cdot \frac {\sin(\ln(\cos(\pi/x)^{1/2}))} {\ln(\cos(\pi/x)^{1/2}))} \cdot \frac {1/4 \,\ln(1 - (\sin(\pi/x)^2))} {(\sin(\pi /x))^2}\cdot \frac{(\sin(\pi/ x))^2}{(\pi/ x)^2}\cdot (\pi/x)^2 $$

Combine the first and last factor to get $\pi^2$, the second and fourth factor both have limit 1, and the middle factor has limit $-1/4$. So the overall answer is $-\pi^2/4$.

Many similar questions can be reduced to fraction algebra and common limits, without using Taylor series.

Answer 4

Notice that

\begin{align*} \ln\big(\cos(\pi/x)^{1/2}\big) &= \frac 1 2 \ln \cos \frac{\pi}{x} \\ &= \frac 1 2 \ln \left(1 - \frac 1 2\frac{\pi^2}{x^2} + O\left(\frac 1 {x^4}\right)\right) \\ &= -\frac 1 4 \frac{\pi^2}{x^2} + O \left( \frac 1 {x^4}\right) \end{align*}

Now do you see how to show that the desired limit is $-\pi^2/4$?

Answer 5

Push $x^2$ to the bottom, then apply L'Hospital rule. $\lim_{x\to \infty} \frac{\sin (\frac{\ln (\cos (\pi/x)}{2})}{x^{-2}}$ \ applying L'Hospital once and some simplification you will get, \ $-\pi^2/4 \lim_{x\to \infty} [\frac{\tan (\pi/x)}{\pi/x} \cos (\frac{\ln (\cos (\pi/x)}{2})]=-\pi^2/4$.

Answer 6

Using $\pi/x=t$, you can compute $$ \lim_{t\to0^+}\frac{\pi^2}{t^2}\sin\left(\frac{1}{2}\ln\cos t\right) $$ Pulling the exponent out of the logarithm is possible because $\cos t$ is positive in a neighborhood of $0$.

Now recall that $$ \cos t=1-\frac{t^2}{2}+o(t^2) $$ so $$ \ln(\cos t)=-\frac{t^2}{2}+o(t^2) $$ and so your limit is $$ \lim_{t\to0^+}-\frac{\pi^2}{4}\frac{t^2+o(t^2)}{t^2} $$

Answer 7

By Taylor limited to the useful terms ($\cos(t)\approx1-\dfrac{t^2}2$, $\ln(1+t)\approx t$, $\sin(t)\approx t$),

$$\cos\left(\frac\pi x\right)\approx1-\frac{\pi^2}{2x^2},$$

$$\ln\left(\left(1-\frac{\pi^2}{2x^2}\right)^{1/2}\right)\approx-\frac12\frac{\pi^2}{2x^2},$$

$$-\sin\left(\frac{\pi^2}{4x^2}\right)\approx-\frac{\pi^2}{4x^2}.$$

Hence $$-\frac{\pi^2}4.$$

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But how do we known how many terms to keep ?

The cosine gives even terms, with the constant coefficient $1$. The logarithm suppresses the coefficient $1$ and leaves the next term unchanged. The sine also leaves this term unchanged.