$$\lim_{x\to 0} {\ln(\cos x)\over \sin^2x} = ?$$

I can solve this by using L'Hopital's rule but how would I do this without this?


closed as off-topic by heropup, Mike Haskel, Henrik, Brevan Ellefsen, Leucippus Aug 16 '16 at 3:08

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  • 2
    $\begingroup$ Do you know Taylor expansions? If yes, use them! $\endgroup$ – paf Aug 15 '16 at 16:46
  • $\begingroup$ I know expansions of cosx , sinx , ln(1+x)...but how to use them here? $\endgroup$ – Sujan Dutta Aug 15 '16 at 16:56
  • $\begingroup$ See my edit below $\endgroup$ – paf Aug 15 '16 at 17:03
  • 1
    $\begingroup$ Basically, there are many many ways to do it :P $\endgroup$ – Patrick Stevens Aug 15 '16 at 17:06
  • $\begingroup$ @PatrickStevens The best and worst parts of Calculus in a nutshell. $\endgroup$ – Simply Beautiful Art Aug 15 '16 at 21:48

$$\frac{\log\left(\cos\left(x\right)\right)}{\sin^{2}\left(x\right)}=\frac{1}{2}\frac{\log\left(1-\sin^{2}\left(x\right)\right)}{\sin^{2}\left(x\right)}=-\frac{\sin^{2}\left(x\right)+O\left(\sin^{4}\left(x\right)\right) }{2\sin^{2}\left(x\right)}\stackrel{x\rightarrow0}{\rightarrow}-\frac{1}{2}.$$

  • $\begingroup$ I was looking for this type of answer. $\endgroup$ – Sujan Dutta Aug 15 '16 at 16:54

The expression equals

$$\frac{\ln (\cos x) - \ln (\cos 0)}{\cos x - \cos 0}\cdot\frac{\cos x - 1}{x^2}\cdot \frac{x^2}{\sin^2 x}.$$

The first fraction $\to \ln'(1) = 1,$ by definition of the derivative. The limit of the second fraction is standard and equals $-1/2.$ The third fraction $\to 1.$ So the limit is $-1/2.$


You can use the important limits: $\lim_{x\rightarrow 0}\frac{\sin x}{x} =1$ and $\lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=e$ (i.e., $\lim_{x\rightarrow 0} \frac{\ln(1+x)}{x}=1$). Then \begin{align*} \lim_{x\rightarrow 0}\frac{\ln \cos x}{\sin^2 x} &= \lim_{x\rightarrow 0}\frac{\cos x -1}{x^2}\\ &=\lim_{x\rightarrow 0}\frac{-2\sin^2 \frac{x}{2}}{x^2}\\ &=-\frac{1}{2}. \end{align*}

  • 1
    $\begingroup$ Need $\neq$ can. But yes, this is the simplest approach (although the OP may not be familiar with the notations. $\endgroup$ – Clement C. Aug 15 '16 at 16:53

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

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

for $x>0$.

Then, we have from $(1)$

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

Now, using the trigonometric identity $\cos(x)-1=-2\sin^2(x/2)$ in $(2)$ reveals

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

whereby application of the squeeze theorem to $(3)$ gives the coveted limit

$$\lim_{x\to 0}\frac{\log(\cos(x))}{\sin^2(x)}=-\frac12$$

Note that we used the result $\log_{x\to 0}\frac{\sin(x)}{x}=1$, which can be obtained a number of ways including using the well-known inequalities from elementary geometry

$$x\cos(x)\le \sin(x)\le x$$

for $|x|\le \pi/2$.


Note that $$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\lim_{x\to 0}\frac{2\sin^2(\frac x2)}{x^2}=\lim_{x\to 0}\frac 14 \frac{2\sin^2(\frac x2)}{\frac{x^2}{4}}=\frac{1}{2}$$ Now, $$\lim_{u\to 1}\frac{\ln u}{u-1}=\ln'(1)=1$$ Hence, since $u=\cos x$ tends to 1 when $x$ tends to 0: $$\lim_{x\to 0}\frac{\ln(\cos x)}{\cos x-1}=1$$ So $$\lim_{x\to 0}\frac{\ln(\cos x)}{x^2} =\lim_{x\to 0}\frac{\ln(\cos x)}{\cos x-1}\lim_{x\to 0}\frac{\cos x-1}{x^2}=-\frac{1}{2}$$

Finally, since $$\lim_{x\to 0} \left(\frac{\sin x}{x}\right)^2 = 1^2=1$$ ,we have $$\lim_{x\to 0}\frac{\ln(\cos x)}{\sin^2x}=\lim_{x\to 0}\frac{\ln(\cos x)}{x^2}\lim_{x\to 0} \left(\frac{x}{\sin x}\right)^2=-\frac{1}{2}$$

Edit: with Taylor expansions, life becomes easier! Near 0, we have $$\frac{\ln(\cos x)}{\sin^2x}=\frac{\ln\left(1-\frac{x^2}{2}+o(x^2)\right)}{x^2+o(x^2)}$$

Since $\ln(1+u)=u+o(u)$ near 0, we obtain (letting $u=-\frac{x^2}{2}+o(x^2)$ above): $$\frac{\ln(\cos x)}{\sin^2x}=\frac{-\frac{x^2}{2}+o(x^2)}{x^2+o(x^2)}$$ near 0. So the limit is $-\frac 12$.


This way doesn't require fiddling with Taylor series or interchanging any sums and limits; it's an example of one of the many places where we can simplify things by recognising a derivative.

Substituting $u = \cos(x)$, we obtain $$\lim_{u \to 1} \frac{\log u}{1-u^2} = \lim_{u \to 1} \left[ \frac{\log u}{1-u} \times \frac{1}{1+u} \right]$$

Now, this is just $$\frac{1}{2} \times \lim_{u \to 1} \frac{\log(u)}{1-u} = -\frac{1}{2} \lim_{h \to 0} \frac{\log(1+h) - \log(1)}{h}$$ where we have substituted $h=-(1-u)$ and introduced the term $\log(1) = 0$ in the numerator.

That final limit is just $\frac{d}{dx} \log(x)$ evaluated at $x=1$; i.e. $1$.

Strictly speaking, I suppose what we should do is find the limit when approached from above, and then the limit when approached from below, and show that they are the same. Otherwise the $u=\cos(x)$ substitution isn't obviously kosher. The calculations will be exactly the same.


$$\frac{\log\left(\cos\left(x\right)\right)}{\sin^{2}\left(x\right)}=\frac{\log\left(1+\cos\left(x\right) -1 \right)}{\sin^{2}\left(x\right)}$$ $$\lim_{x\to 0} \frac{\log\left(1+\cos\left(x\right) -1 \right)}{\sin^{2}\left(x\right)}.\frac{x^2}{x^2}.\frac{\left(\cos\left(x\right) -1 \right)}{\left(\cos\left(x\right) -1 \right)}$$

  • $\begingroup$ Hm, this answer feels somewhere between incomplete and a duplicate of zhw's answer. $\endgroup$ – Simply Beautiful Art Aug 15 '16 at 21:55
  • $\begingroup$ @SimpleArt What is incomplete ? $\endgroup$ – Aakash Kumar Aug 16 '16 at 1:05
  • $\begingroup$ It is simply not as enlightening as zhw's answer and requires too much effort figuring out how to arrange the terms to get the right form for a good limit. (at least in my opinion, but I didn't downvote simply because of this) $\endgroup$ – Simply Beautiful Art Aug 16 '16 at 1:08
  • $\begingroup$ @SimpleArt I don't think it require too much effort , especially to those who are good at limit .The whole question can be easily solved in two steps . $\endgroup$ – Aakash Kumar Aug 16 '16 at 3:38

We can solve this problem by using our knowledge of limits for composite functions.

Let $u = \sin^2 x.$

$x \to 0 \implies \sin^2 x \to 0 \implies u \to 0$

$\frac{\ln (\cos x)}{\sin^2x} = \frac{\ln (\cos^2 x)}{2\sin^2x} = \frac{\ln (1 - \sin^2 x)}{2\sin^2x} = \frac{\ln (1 - u)}{2u} = 1/2 \ln(1-u)^\frac{1}{u}$

Note that $\ln(1-u)^\frac{1}{u}$ is continuous at u = 0 and is equal to $1/e$, so we use the limit rule for composite functions to get

$\lim_{x \to 0} \frac{\ln (\cos x)}{\sin^2 x} = \frac{1}{2} \lim_{u \to 0}\ln(1-u)^\frac{1}{u} = \frac{1}{2}\ln \frac{1}{e} = -\frac{1}{2}$


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