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.
 A: 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.
A: 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 $\displaystyle \lim_{x \to 0+} \ln(\ell) = \ln(1) =0$.
A: 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 $0$.
A: 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 $\displaystyle x \to 0$.
A: $$\lim_{x \to 0} \dfrac{\ln(\cos(x))}{x} = \lim_{x \to 0} \dfrac{\ln(1-2\sin^2(x/2))}{x} = \lim_{x \to 0} \dfrac{\ln(1-2\sin^2(x/2))}{2 \sin^2(x/2)} \dfrac{\sin^2(x/2)}{x/2} = \lim_{x \to 0} \dfrac{\ln(1-2\sin^2(x/2))}{2 \sin^2(x/2)} \lim_{x \to 0} \dfrac{\sin(x/2)}{x/2} \lim_{x \to 0} \sin(x/2) = -1 \times 1 \times 0 = 0$$
A: We can write
$$L=\lim\limits_{x \to 0^+} \frac{\ln[\cos(x)]}{x}=\lim_{x \to 0^+}\frac{\ln[1+(\cos x -1)]}{x}$$
Then clearly as $x \to 0^+$ , $$L \sim \frac{\cos x -1}{x} \sim \frac{-x}{2} \to 0$$
