# Solving $\lim\limits_{x \to 0^+} \frac{\ln[\cos(x)]}{x}$

I had a test today which involved solving the following limit:

$$\lim\limits_{x \to 0^+} \frac{\ln[\cos(x)]}{x}$$

I didn't figure out how to solve. After the test, I asked a couple of classmates and they told me that it was supposed to be solved by first transporting the multiplication by $\frac{1}{x}$ inside the logarithm as an exponent and then replacing $\cos(x)$ with $\sqrt{1-\sin^2(x)}$, giving the following expression:

$$\lim\limits_{x \to 0^+} \ln[(1-\sin^2(x))^\frac{1}{2x}]$$

However, I don't know how to proceed from here. It's not like it matters a lot at this point, but I'd still like to know how I'm supposed to solve this. I'm not sure if it's applicable here, but we haven't learned L'Hôpital's rule.

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Note that $\ln(\cos(0))=0$.

So we can write our limit as $$\lim_{x\to 0^+} \frac{\ln(\cos(x))-\ln(\cos(0))}{(x-0)}.$$

Note that the above expression is almost the usual expression for the derivative of $\ln(\cos(x))$ at $x=0$. (If necessary, go back and look up the definition of the derivative of $f(x)$ at $x=a$). The only difference is the use of a one-sided limit. But what about if the two-sided limit existed?

If we are very lucky and the derivative of $\ln(\cos(x))$ at $0$ exists, the value of that derivative at $x=0$ will be our answer.

So differentiate $\ln(\cos(x))$ in the usual way. Everything works out nicely, the derivative is $0$.

Added: I expect there is no issue in finding the derivative, but here are the details. Using the Chain Rule, we get $$-(\sin(x))\frac{1}{\cos(x)}.$$ At $x=0$ this is $0$.

So the $0^+$ turns out to be unnecessary, plain old $0$ will do. The manipulations suggested by classmates are not needed, everything follows from the definition of derivative, if we know a couple of differentiation rules.

Comment: This is not really how I would do it, if I needed to know the answer. The "natural" approach is to use the power series expansions of $\cos(x)$ and $\ln(1+u)$. But since you mentioned that you had not yet done L'Hospital's Rule, I assumed that you would not yet have been exposed to power series.

But the power series approach is very much worth knowing. The power series expansion of $\cos x$ is $$1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!} +\cdots.$$

So very informally, if $x$ is near $0$, $\cos x$ is about $1-x^2/2$.

The power series expansion of $\ln(1+u)$ is $$u-\frac{u^2}{2}+\frac{u^3}{3} -\frac{u^4}{4}+\cdots.$$ (This expansion is only valid when $-1 \lt u \le 1$.)

So when $x$ is near $0$, $\ln(\cos(x))$ is about $-x^2/2$. Divide by $x$. We get $-x/2$, which approaches $0$ as $x$ approaches $0$.

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Forgive me if I'm missing something, but what does the derivative have to do with this? –  Javier Badia Jul 16 '11 at 1:50
@Javier: what's the definition of the derivative? –  lentic catachresis Jul 16 '11 at 1:53
@Bruno: So what this is doing is sort of taking the expression and trying to transform it into something that will fit the definition of the derivative and then work from there? –  Javier Badia Jul 16 '11 at 1:56
@Javier: Exactly! You try to see your limit as the derivative of a function on a given point. Once you got it that way, you calculate the derivative of the recognized function using standard methods, evaluate it on the point, and voilà. –  lentic catachresis Jul 16 '11 at 1:59
@Javier Badia: There is seldom a "best" answer. As I pointed out, in real life I would use series, to get a handle on the error term. However, exams are not real life. If this was on a calculus exam or problem set, after derivatives but before series, the derivative approach is undoubtedly what you were expected to use. If the problem was medium early in an analysis course, something like what Aryabhata did would be the expected kind of approach. –  André Nicolas Jul 16 '11 at 14:38

We can also do this without the knowledge of derivatives (which I presume you haven't yet gotten to).

We know that $\displaystyle \lim_{x \to 0} (1-x)^{1/x} = e^{-1}$. Thus for $x$ sufficiently close to $\displaystyle 0$

we have that $\displaystyle (1-x)^{1/x} \gt e^{-2}$ and thus taking logs, we get

$$\frac{\ln (1-x)}{x} \ge -2$$ and so

$$\ln(1-x) \ge -2x$$ for $\displaystyle x$ sufficiently close to $\displaystyle 0$.

Thus

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

Thus for $\displaystyle x$ sufficiently close to $\displaystyle 0$ we have that

$$0 \ge \frac{\ln (\cos x)}{x} \ge \frac{-\sin^2 x}{x}$$

Since we know that $\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1$, by the sandwich theorem we have that the limit you seek is $\displaystyle 0$, as $\displaystyle \frac{\sin^2 x}{x} \to 0$ as $\displaystyle x \to 0$.

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This seems to make sense, but why did you write $\frac{\ln(1-\cos x)}{x}$ instead of $\frac{\ln(\cos x)}{x}$ in the last step, and how do you know that it's less than or equal to 0? –  Javier Badia Jul 16 '11 at 2:39
@Javier: That is a typo :) And $\ln t \le 0$ for $0 \lt t \lt 1$, –  Aryabhata Jul 16 '11 at 2:41
Of course, should have realized that. Thanks. –  Javier Badia Jul 16 '11 at 2:45

HINT $\$ Many limits can be simply calculated by recognizing them as instances of first derivatives and then calculating the derivatively rotely using known derivative rules. You can find a handful of examples in my prior posts starting here.

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this is a good point to know (+1). –  Chris's sis Jun 20 '12 at 18:06

Let \begin{align*} \ell = \lim_{x \to 0+} (\cos{x})^{\frac{1}{x}} &= \lim_{x \to 0+} (1+ (\cos{x}-1))^{\frac{1}{x}} \\&= \lim_{x \to 0+}(1+(\cos{x}-1))^{\frac{1}{\cos{x}-1} \times \frac{\cos{x}-1}{x}} \\ &= e^{\scriptsize{\displaystyle\lim_{x \to 0+} \frac{\cos{x}-1}{x}}} \\ &= e^{0} =1 \qquad\qquad\Bigl[ \because \small\lim_{x \to 0} \frac{\cos{x}-1}{x} = \lim_{x \to 0}\frac{-2\:\sin^{2}\frac{x}{2}}{x} = -\:\lim_{x \to 0} \frac{\sin\frac{x}{2}}{\frac{x}{2}} \cdot \sin\frac{x}{2} \Bigr] \end{align*}

Hence $\displaystyle \lim_{x \to 0+} \ln(\ell) = \ln(1) =0$.

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$$\lim_{x \to 0} \dfrac{\ln(\cos(x))}{x} = \lim_{x \to 0} \dfrac{\ln(1-2\sin^2(x/2))}{x} = \lim_{x \to 0} \dfrac{\ln(1-2\sin^2(x/2))}{2 \sin^2(x/2)} \dfrac{\sin^2(x/2)}{x/2} = \lim_{x \to 0} \dfrac{\ln(1-2\sin^2(x/2))}{2 \sin^2(x/2)} \lim_{x \to 0} \dfrac{\sin(x/2)}{x/2} \lim_{x \to 0} \sin(x/2) = -1 \times 1 \times 0 = 0$$

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