Question

$$\lim_{x \to (\frac{1}{2})^{-}} \frac{\ln(1 - 2x)}{\tan \pi x}$$

I'm not sure how to go about this limit, I've tried to apply L'Hopital's rule (as shown).

It seems that the form is going to be forever indeterminate? Unless I'm missing something, but the limit is (apparently) zero.

Working

As this is in the indeterminate form of $- \infty / + \infty$, apply L'Hopital's rule.

Let $\frac{\ln(1 - 2x)}{\tan \pi x} = f/g$, then

$f' = \frac{-2}{1 - 2x}$ , and $f'' = \frac{-4}{(1 - 2x)^2}$

$g' = \pi \sec^2 \pi x$ (which is equal to $\pi (\tan^2 (\pi x )+ 1)$ ) and

$g'' = 2 \pi^2 \sec^2 (\pi x) \tan \pi x$ ( or $2 \pi^2 (\tan^2 (\pi x) + 1) \tan \pi x$ )

Using the first derivatives gives

$$\lim_{x \to (\frac{1}{2})^{-}} \frac{\frac{-2}{1 - 2x}}{\pi \sec^2 \pi x}$$

Which is in the form

$$\frac{- \infty}{+ \infty}$$

So that I would now use the second derivative, which is

$$\lim_{x \to (\frac{1}{2})^{-}} \frac{ \frac{-4}{(1 - 2x)^2} }{ 2 \pi^2 \sec^2 (\pi x) \tan \pi x }$$

Or

$$\lim_{x \to (\frac{1}{2})^{-}} \frac{ \frac{-4}{(1 - 2x)^2} }{ 2 \pi^2 (\tan^2 (\pi x) + 1) \tan \pi x }$$

But this is still in an indeterminate form?

You were doing fine:

Using the first derivatives gives $$\lim_{x \to (\frac{1}{2})^{-}} \frac{\frac{-2}{1 - 2x}}{\pi \sec^2 \pi x}$$

Now simplify a bit first: $$\frac{\frac{-2}{1 - 2x}}{\pi \sec^2 \pi x} = \frac{2\cos^2\pi x}{\pi \left(2x-1\right)}$$ Applying l'Hôpital to the fraction in this form gives the easy $-\sin\left( 2\pi x\right)$, which evaluates to $0$.

• Thanks, this was easiest for me to follow, and probably what the text may expected. Some notes to consider adding, maybe they're useful to others learning. ($*$) As $D_x \left( 2 \cos^2 \pi x \right) = -4 \cos(\pi x) \sin (\pi x)$ , using trig identity $2\sin(a)\cos(a) = sin(2a)$ gives $-4 \cos(\pi x) \sin (\pi x) = - 2 \sin (2 \pi x)$. Then $D_x(\pi (2x - 1)) = 2 \pi$. We then have $$\frac{- 2 \sin (2 \pi x)}{2 \pi }= \frac{ - \sin (2 \pi x) }{ \pi}$$ And as $x \to 1/2$ (from $x < 1/2$ ) then $\sin(2 \pi x) \to 0$, which gives the result $\frac{0}{\pi} = 0$. – baxx Jul 19 '17 at 13:03
• @baxx You're welcome! Don't forget the extra factor $\pi$ from the chain rule, which is why it even simplifies to $-\sin 2\pi x$. – StackTD Jul 19 '17 at 13:05
• ah yes >.<, I was write by "luck" rather than method there, as the denominator doesn't affect the value as it's constant (and the numerator $\to 0$. – baxx Jul 19 '17 at 13:09
• Yes, the result happens to stay the same here but your teacher might not like forgetting about the chain rule ;-). – StackTD Jul 19 '17 at 13:10

As $x<\dfrac12\iff2x-1<0$

set $1-2x=y$

$$F=\lim_{y\to0^+}\dfrac{\ln y}{\tan\dfrac{\pi(1-y)}2}=\lim_{y\to0^+}\dfrac{\ln y}{\cot\dfrac{\pi y}2}$$

Applying L'Hospital's $$F=\lim_{y\to0^+}\dfrac{\sin^2\dfrac{\pi y}2}{-\dfrac\pi2\cdot y}\cdots=0$$ using $\lim_{h\to0}\dfrac{\sin h}h=1$

Let $y = \displaystyle \frac{1}{2} - x$. As $x \to \displaystyle \Big(\frac{1}{2}\Big)^{-}$, $y \to 0^{+}$. Then:

$$\lim \limits_{x \to \Big( \frac{1}{2} \Big)^{-}} \frac{\ln(1-2x)}{\tan(\pi x)} = \lim \limits_{x \to 0^{+}} \frac{\ln(2y)}{\tan\Big( \frac{\pi}{2} - \pi y \Big)}$$

Because $\displaystyle \tan\Big( \frac{\pi}{2} - \pi y \Big) = \frac{1}{\tan(\pi y)}$, we have:

$$\lim \limits_{x \to 0^{+}} \frac{\ln(2y)}{\tan\Big( \frac{\pi}{2} - \pi y \Big)} = \lim \limits_{x \to 0^{+}} \ln(2y) \tan(\pi y).$$

As $y \to 0$, $\tan(\pi y) \sim \pi y$. Therefore:

$$\ln(2y) \tan(\pi y) \mathop{\sim} \limits_{y \to 0^{+}} \pi y \ln(2y).$$

As a consequence :

$$\lim \limits_{x \to \Big( \frac{1}{2} \Big)^{-}} \frac{\ln(1-2x)}{\tan(\pi x)} = 0.$$

Let $x-\frac{1}{2}=y$. Hence, $x=\frac{1}{2}+y$ and $$\lim_{x \to \frac{1}{2}^{-}} \frac{\ln(1 - 2x)}{\tan \pi x} =-\lim_{y\rightarrow0^-}\frac{\ln(-y)}{\cot{\pi y}}=\pi\lim_{y\rightarrow0^-}\left(-y\ln(-y)\right)=0$$

You do not need de l'Hospital rule for the evaluation of such limit: $$\lim_{x\to(1/2)^{-}}\frac{\log(1-2x)}{\tan(\pi x)}=\lim_{z\to 0^+}\frac{\log(2z)}{\cot(\pi z)}=\lim_{z\to 0^+}\frac{z\log(2z)}{z\cot(\pi z)} =\frac{0}{\frac{1}{\pi}}=0.$$ It is enough to exploit a substitution $x\mapsto \frac{1}{2}-z$ and the well-known limits (that should be studied before approaching de l'Hospital rule, the Stolz-Cesàro theorem and so on) $$\lim_{x\to 0}\frac{\sin x}{x}=1,\qquad \forall\epsilon>0,\;\;\lim_{x\to 0^+} x^{\varepsilon}\log(x)=0.$$

• It seems that more things need to be known in this approach, but it's nice to see one without the rule. – baxx Jul 19 '17 at 13:11
• @baxx: it is just a substitution $x\mapsto \frac{1}{2}-z$ and two well-known limits... – Jack D'Aurizio Jul 19 '17 at 13:12
• I'm not familiar with them, others might not be either. – baxx Jul 19 '17 at 13:15
• I wonder what teachers teach these days. De l'Hopital's rule before proving $$\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{e^x-1}{x}=\lim_{x\to 0}\frac{\log(1+x)}{x}=1, \qquad \lim_{x\to 0^+} x^{\varepsilon}\log(x) = 0$$ ? Sounds strange to me. – Jack D'Aurizio Jul 19 '17 at 13:25
• I think that the identities used should be mentioned in an answer , you're welcome to lament about the education system though if you like – baxx Jul 19 '17 at 13:27