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7.5 I · Q66

Q.lim⁡x→1[1−x2sin⁡πx]\displaystyle\lim_{x\to 1}\left[\frac{1-x^2}{\sin \pi x}\right]

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Let t=x−1t=x-1, so x=1+tx=1+t and t→0t\to0. 1−x2=1−(1+t)2=−2t−t2=−t(2+t)1-x^2=1-(1+t)^2=-2t-t^2=-t(2+t). Also sin⁡(πx)=sin⁡(π+πt)=−sin⁡(πt)\sin(\pi x)=\sin(\pi+\pi t)=-\sin(\pi t). So the expression is $\dfrac{-t(2+t)}{-\sin(\pi t)}=\dfrac{t(2+t)}{\sin(\pi t)}=\dfrac{(2+t)}{\pi} …

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