Question Number 122548 by bramlexs22 last updated on 18/Nov/20
$$\:\:\underset{{x}\rightarrow\mathrm{1}/\mathrm{2}} {\mathrm{lim}}\:\left(\frac{\mathrm{cot}\:\left(\pi{x}\right)}{\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}}\right)=?\: \\ $$
Answered by liberty last updated on 18/Nov/20
![lim_(x→1/2) ((cot (πx))/((2x−1)(x+1))) = lim_(x→1/2) (1/(x+1)).lim_((((2x−1)/2))→0) ((cot (πx))/((2x−1))) = (2/3) × lim_(X→0) ((cot (π(((2X+1)/2))))/(2X)) ; [ let ((2x−1)/2)=X ] = (1/3) × lim_(X→0) ((cot ((π/2)+πX))/X) = (1/3) × lim_(X→0) ((−tan (πX))/X) = −(π/3).▲](https://www.tinkutara.com/question/Q122549.png)
$$\:\underset{{x}\rightarrow\mathrm{1}/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{cot}\:\left(\pi\mathrm{x}\right)}{\left(\mathrm{2x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)}\:=\:\underset{{x}\rightarrow\mathrm{1}/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}.\underset{\left(\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{2}}\right)\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cot}\:\left(\pi\mathrm{x}\right)}{\left(\mathrm{2x}−\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\:×\:\underset{\mathrm{X}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cot}\:\left(\pi\left(\frac{\mathrm{2X}+\mathrm{1}}{\mathrm{2}}\right)\right)}{\mathrm{2X}}\:\:\:\:\:\:\:;\:\left[\:\mathrm{let}\:\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{2}}=\mathrm{X}\:\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\:×\:\underset{\mathrm{X}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cot}\:\left(\frac{\pi}{\mathrm{2}}+\pi\mathrm{X}\right)}{\mathrm{X}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\:×\:\underset{\mathrm{X}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{tan}\:\left(\pi\mathrm{X}\right)}{\mathrm{X}}\:=\:−\frac{\pi}{\mathrm{3}}.\blacktriangle \\ $$
Answered by Dwaipayan Shikari last updated on 18/Nov/20
$$\underset{{x}\rightarrow\frac{\mathrm{1}}{\mathrm{2}}} {\mathrm{lim}}\frac{{cos}\left(\pi{x}\right)}{{sin}\left(\pi{x}\right)\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)} \\ $$$$\underset{{x}\rightarrow\frac{\mathrm{1}}{\mathrm{2}}} {\mathrm{lim}}\frac{\mathrm{2}}{\mathrm{3}}\:\frac{{sin}\left(\frac{\pi}{\mathrm{2}}−\pi{x}\right)}{\left(\mathrm{2}{x}−\mathrm{1}\right)}=\frac{\mathrm{2}}{\mathrm{3}}.\frac{\pi\left(\mathrm{1}−\mathrm{2}{x}\right)}{\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}\right)}=−\frac{\pi}{\mathrm{3}} \\ $$