Question Number 125790 by mohammad17 last updated on 13/Dec/20
Answered by liberty last updated on 13/Dec/20
![lim_(x→2) (((x+2)(x−2))/(sin πx)) = 4×lim_(x→2) [ ((x−2)/(sin πx)) ] [ let x=2+t ∧ t→0 ] = 4× lim_(t→0) (t/(sin π(t+2))) = 4 ×lim_(t→0) (t/(sin (2π+πt))) = 4 × lim_(t→0) (t/(sin πt)) = 4× lim_(t→0) ((πt)/(πsin πt)) = (4/π) × lim_(t→0) ((πt)/(sin πt)) = (4/π)×1=(4/π)](Q125792.png)
$$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\left({x}+\mathrm{2}\right)\left({x}−\mathrm{2}\right)}{\mathrm{sin}\:\pi{x}}\:=\:\mathrm{4}×\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\left[\:\frac{{x}−\mathrm{2}}{\mathrm{sin}\:\pi{x}}\:\right] \\ $$$$\:\left[\:{let}\:{x}=\mathrm{2}+{t}\:\wedge\:{t}\rightarrow\mathrm{0}\:\right]\: \\ $$$$=\:\mathrm{4}×\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{t}}{\mathrm{sin}\:\pi\left({t}+\mathrm{2}\right)}\:=\:\mathrm{4}\:×\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{t}}{\mathrm{sin}\:\left(\mathrm{2}\pi+\pi{t}\right)} \\ $$$$=\:\mathrm{4}\:×\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{t}}{\mathrm{sin}\:\pi{t}}\:=\:\mathrm{4}×\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\pi{t}}{\pi\mathrm{sin}\:\pi{t}} \\ $$$$=\:\frac{\mathrm{4}}{\pi}\:×\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\pi{t}}{\mathrm{sin}\:\pi{t}}\:=\:\frac{\mathrm{4}}{\pi}×\mathrm{1}=\frac{\mathrm{4}}{\pi} \\ $$
Commented by mohammad17 last updated on 13/Dec/20
$${nise}\:{sir}\:{thank}\:{you} \\ $$
Answered by mathmax by abdo last updated on 14/Dec/20
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{4}}{\mathrm{sin}\left(\pi\mathrm{x}\right)}\:\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}+\mathrm{2}\right)}{\mathrm{sin}\left(\pi\mathrm{x}\right)}=_{\mathrm{x}−\mathrm{2}=\mathrm{t}} \:\:\:\frac{\mathrm{t}\left(\mathrm{t}+\mathrm{4}\right)}{\mathrm{sin}\left(\pi\left(\mathrm{t}+\mathrm{2}\right)\right)} \\ $$$$=\frac{\mathrm{t}\left(\mathrm{t}+\mathrm{4}\right)}{\mathrm{sn}\left(\pi\mathrm{t}\right)}\:\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{x}\rightarrow\mathrm{2}\:\Leftrightarrow\mathrm{t}\rightarrow\mathrm{0}\:\Rightarrow\frac{\mathrm{t}\left(\mathrm{t}+\mathrm{4}\right)}{\mathrm{sin}\left(\pi\mathrm{t}\right)}=\sim\frac{\mathrm{t}\left(\mathrm{t}+\mathrm{4}\right)}{\pi\mathrm{t}}\sim\frac{\mathrm{t}+\mathrm{4}}{\pi}\rightarrow\frac{\mathrm{4}}{\pi} \\ $$$$\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{4}}{\pi} \\ $$
Answered by Dwaipayan Shikari last updated on 14/Dec/20
$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{sin}\pi{x}}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}−\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{sin}\left(\mathrm{2}\pi−\pi{x}\right)}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}−\frac{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)}{\pi\left(\mathrm{2}−{x}\right)}=\frac{{x}+\mathrm{2}}{\pi}=\frac{\mathrm{4}}{\pi} \\ $$