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$$\lim_{x\to 1}\ln(1-x)\cot\frac{\pi x}2$$ After applying L'Hospital twice, I get $$\lim_{x\to 1}\frac{-2\sin\pi x}\pi = 0$$

Is this correct? And if I do by LHL and RHL method, ln(1-x) would not be defined for RHL since the log of negative is not defined. Also the concern that tan($\pi$/2) approaches +$\infty$ from LHL and -$\infty$ from RHL.

T.Pal
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  • $\text{cotan}(\pi x/2)$ is $\sim x-1$ near 1 which goes much stronger to zero then the log diverges so i think ur result is correct – tired Mar 28 '16 at 12:13
  • Thanks mate. :) Any counter arguments would be much appreciated from people on stackexchange.. – T.Pal Mar 28 '16 at 12:19

3 Answers3

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$$\lim_{x\to1^-}\frac{\log(1-x)\cos\frac{\pi x}2}{\sin\frac{\pi x}2}=\lim_{x\to1^-}\frac1{\sin\frac{\pi x}2}\cdot\lim_{x\to1^-}\frac{\log(1-x)}{\frac{1}{\cos\frac{\pi x}2}}\stackrel{\text{l'H}}=$$

$$=1\cdot\lim_{x\to1^-}\frac2\pi\frac{-\frac1{1-x}}{-\frac{\sin\frac{\pi x}2}{\cos^2\frac{\pi x}2}}=\frac2\pi\lim_{x\to1^-}\frac1{\sin\frac{x\pi}2}\lim_{x\to1^-}\frac{\cos^2\frac{\pi x}2}{1-x}\stackrel{\text{l'H}}=1\cdot\lim_{x\to1^-}\frac{-2\cos\frac{\pi x}2\sin\frac{\pi x}2}{-1}=0$$

So yes: your result is correct but only as one-sided limit, as the function's isn't defined on any right neighborhood of $\;1\;$ (the splitting of limits is justified since each limit by itself exists finitely)

DonAntonio
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  • But for x approaching from 1- i.e. LHL, the log would not be defined. Can we still claim that since RHL is 0 and LHL is not defined, the result would be 0? – T.Pal Mar 28 '16 at 12:28
  • @T.Pal No, we can't. The limit cannot exist since the function isn't even defined on any right neighbourhood of $;1;$, so all we can say is the left limit at $;1;$ exists and vanishes. BTW, I confused between $;+;$ and $;-;$ in the limits. It is corrected now – DonAntonio Mar 28 '16 at 12:32
  • Yes that's my concern. So essentially, the limit does not exist as it only exists for one side and not the other and hence no question of equality between LHL and RHL. Is that it? – T.Pal Mar 28 '16 at 12:44
  • @T.Pal Indeed so. – DonAntonio Mar 28 '16 at 12:54
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The function is not defined for $x>1$; in this case writing $\lim_{x\to1}$ or $\lim_{x\to1^-}$ is just a question of conventions.

A possible simplification is to set $1-x=\frac{2t}{\pi}$, so we have $$ \cot\frac{\pi x}{2}= \cot\left(\frac{\pi}{2}-t\right)= \tan t $$ and the limit becomes $$ \lim_{t\to0^+}\ln\frac{2t}{\pi}\tan t= \lim_{t\to0^+}\left(\ln\frac{2}{\pi}\tan t+\ln t\tan t\right)=0 $$ because the first summand has limit $0$ and the second can be written $$ \lim_{t\to0^+}(t\ln t)\frac{\tan t}{t}=0\cdot 1=0 $$

egreg
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  • Interesting. This proves that the limit indeed is zero. Also I understand that it was a mere formality to prove the RHL since the function is not even defined for x>1 and I could have concluded that the limit is ZERO without carrying out the RHL proof explicitly. Correct me if I am wrong here. – T.Pal Mar 28 '16 at 13:04
  • @T.Pal The limit from the right is meaningless. – egreg Mar 28 '16 at 13:06
  • But then your proof works for Right limit and shows it is zero, albeit using a substitution to simplify. – T.Pal Mar 28 '16 at 13:08
  • @T.Pal If $x<1$, then under the substitution we have $t>0$, so $\lim_{x\to1^-}$ becomes $\lim_{t\to0^+}$ – egreg Mar 28 '16 at 13:09
  • Missed that. So only one sided limit exists as the other side is meaningless in R. Thanks. :) – T.Pal Mar 28 '16 at 13:12
-1

Lim log(1-x) cot(pi x/2)

x->1

Let x-1=t

Lim log t cot((pi/2)(t+1))

t->0

Lim log t cot((pi/2) +(pi t/2))

t->0

Lim log t tan(pi t/2)

t->0

Lim log t /cot(pi t/2)

t->0

On putting limit, we get infinity/infinity form. Therefore we can apply L 'Hospital rule

Lim -1/[(pi/2)cosec²(pi t/2)]

t->0

Lim sin²(pi t/2)/(pi t/2)

t->0

Multiplying and dividing by pi t/2

Lim (sin²(pi t/2) / (pi t/2)² )× pi t/2

t->0

Using lim x->0 sin²x/x² =1

Lim pi t/2

t->0

=0

user1001118
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  • Welcome to MSE. When you post a question with a lot of math typing, maybe try using https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference. – Tereza Tizkova Dec 05 '21 at 11:06