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Question Number 193532 by horsebrand11 last updated on 16/Jun/23

lim_(x→2) ((1−cos πx)/((2−x)^2 )) =?

$$\:\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\pi\mathrm{x}}{\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{2}} }\:=? \\ $$

Answered by aba last updated on 16/Jun/23

let t=2−x lim_(x→2) ((1−cosπx)/((2−x)^2 ))=lim_(t→0) ((1−cosπ(2+t))/t^2 )=lim_(t→0) ((1−cosπt)/((πt)^2 ))×π^2 =(π^2 /2)

$$\mathrm{let}\:\mathrm{t}=\mathrm{2}−\mathrm{x} \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\pi\mathrm{x}}{\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{2}} }=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\pi\left(\mathrm{2}+\mathrm{t}\right)}{\mathrm{t}^{\mathrm{2}} }=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\pi\mathrm{t}}{\left(\pi\mathrm{t}\right)^{\mathrm{2}} }×\pi^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$

Answered by cortano12 last updated on 16/Jun/23

by L′Hopital L=lim_(x→2) ((πsin πx)/(−2(2−x))) = lim_(x→2) ((π^2 cos πx)/2) = (π^2 /2)

$$\:\mathrm{by}\:\mathrm{L}'\mathrm{Hopital}\: \\ $$$$\:\mathrm{L}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\pi\mathrm{sin}\:\pi\mathrm{x}}{−\mathrm{2}\left(\mathrm{2}−\mathrm{x}\right)}\:=\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\pi^{\mathrm{2}} \mathrm{cos}\:\pi\mathrm{x}}{\mathrm{2}} \\ $$$$\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$

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