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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.

5 Answers 5

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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 $0$.

  • 0
    @EricAm: We are trying to compute a limit. We notice that limit is close to the **definition** of the derivative of a certain function. Suppose we are past the part in beginning calculus where we computed the derivatives of $\sin$, and then $\cos$, and $\log$, and did the Chain Rule. Then we can use these facts to compute limits. There is no circularity involved.2013-10-08
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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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    Of course, should have realized that. Thanks.2011-07-16
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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).2012-06-20
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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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$\lim_{x \to 0} \dfrac{\ln(\cos(x))}{x} = \lim_{x \to 0} \dfrac{\ln(1-2\sin^2(x/2))}{x} = \lim_{x \to 0} \dfrac{\ln(1-2\sin^2(x/2))}{2 \sin^2(x/2)} \dfrac{\sin^2(x/2)}{x/2} = \lim_{x \to 0} \dfrac{\ln(1-2\sin^2(x/2))}{2 \sin^2(x/2)} \lim_{x \to 0} \dfrac{\sin(x/2)}{x/2} \lim_{x \to 0} \sin(x/2) = -1 \times 1 \times 0 = 0$