Q.
f(x)=n→∞lim(x−1)2n+1(x−1)2n−1 is discontinuous at
1804
187
Continuity and Differentiability
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Solution:
f(x)=n→∞lim[(x−1)2]n+1[(x−1)2]n−1 =n→∞lim1+[(x−1)2]n11−[(x−1)2]n1 =⎩⎨⎧−1,0,1,0≤(x−1)2<1(x−1)2=1(x−1)2>1 =⎩⎨⎧1,0,−1,0,1,x<0x=00<x<2x=2x>2
Thus, f(x) is discontinuous at x=0,2