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$$\lim_{x\to 0}\frac{\log\left(\cos x\right)}{x^2}$$

I've been triyng to:

  1. show $\displaystyle -\frac{\pi}{2}<x<\frac{\pi}{2}\Rightarrow\frac{\log\left(\cos x\right)}{x^2}<-\frac{1}{2}$

  2. find a function so that $\displaystyle f(x)<\frac{\log\left(\cos x\right)}{x^2}$ and $\displaystyle \lim\limits_{x\to0}f(x) = -\frac{1}{2}$

And then apply the squeeze principle, but haven't managed any of these.

Michael Rozenberg
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j3M
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6 Answers6

16

HINT:

$$\dfrac{\log(\cos x)}{x^2}=\dfrac{\log(\cos^2x)}{2x^2}=-\dfrac12\cdot\dfrac{\log(1-\sin^2x)}{-\sin^2x}\cdot\left(\dfrac{\sin x}x\right)^2$$

lab bhattacharjee
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4

Use the following:

$$\cos(x)\sim1-\frac12x^2$$

$$\ln(1-x)\sim-x$$

Combine them and you should get

$$\ln(\cos(x))\sim-\frac12x^2$$

$$\lim_{x\to0}\frac{\ln(\cos(x))}{x^2}=-\frac12$$

Simply Beautiful Art
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    Taylor also involves taking derivatives... –  Dec 31 '16 at 16:26
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    One can't "compose" equivalences like that formally. For example, $\sin x \sim x$ but $\exp \left(- \frac{1}{\sin^2 x}\right)$ is not equivalent to $\exp\left(-\frac{1}{x^2}\right)$. The downvote wasn't mine, though. – MathematicsStudent1122 Dec 31 '16 at 16:35
  • @MathematicsStudent1122 That's perfectly fine. A more rigorous approach can be done with bounds and squeeze theorem with bounds on the functions. – Simply Beautiful Art Dec 31 '16 at 16:37
4

I thought it might be instructive to present an approach that relies on a set of elementary inequalities along with the squeeze theorem. To that end, we begin with a short primer.

PRIMER:

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

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

for $x>0$.

And in THIS ANSWER, I developed the well-known inequalities often introduced in geometry

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

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

First, we write $\cos(x)=1-(1-\cos(x))= 1-2\sin^2(x/2)$. Then, using $(1)$ reveals

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

Next, using $(2)$, we obtain

$$\frac{-(x^2/2)}{1-\sin^2(x/2)}\le \log(\cos(x))\le -x^2/2\cos^2(x/2) \tag 4$$

whereupon dividing $(4)$ by $x^2$ yields

$$-\frac12 \frac{1}{1-\sin^2(x/2)}\le \frac{\log(\cos(x))}{x^2}\le -\frac12 \cos^2(x/2)\tag 5$$

Finally, applying the squeeze theorem to $(5)$ yields

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{\log(\cos(x)}{x^2}=-\frac12}$$

And we are done!

Community
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Mark Viola
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3

The limit is the logarithm of

$$\lim_{t\to0}\left(\cos t\right)^{1/t^2}=\lim_{t\to0}\left(1-2\sin^2\frac t2\right)^{1/t^2}=\lim_{t\to0}\left(\left(1-2\sin^2\frac t2\right)^{1/\sin^2(t/2)}\right)^{\sin^2(t/2)/t^2}.$$

The expression inside the outer parenthesis tends to $e^{-2}$, while the exponent is

$$\left(\frac12\frac{\sin\frac t2}{\frac t2}\right)^2$$ which tends to $1/4$.

Hence,

$$-\frac12.$$

3

$$\lim\limits_{x→0}\frac{\log(\cos x)}{x^2}=\lim\limits_{x→0}\frac{\log(1+(\cos x-1))}{x^2}=\lim\limits_{x→0}\frac{\log(1+(\cos x-1))}{\cos x-1}\lim\limits_{x→0}\frac{\cos x-1}{x^2}=\lim\limits_{x→0}\frac{\log(1+(\cos x-1))}{\cos x-1}\lim\limits_{x→0}-\frac{1-\cos x}{x^2}=-\frac{1}{2}$$

Speedding
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$\lim\limits_{x\rightarrow0}\frac{\ln\cos x}{x^2}=\ln\lim\limits_{x\rightarrow0}(1+\cos x-1)^{\frac{1}{\cos x-1}\cdot}\frac{-2\sin^2\frac{x}{2}\cdot}{4\cdot\frac{x^2}{4}}=\ln e^{-\frac{1}{2}}=-\frac{1}{2}$

Michael Rozenberg
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