The value of f(0) such f(x)=1−cos2x+sin2x√x2+1−1(x≠0) is continuous at x=0 is
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Solution
Given f(x)=1−cos2x+sin2x√x2+1−1(x≠0) Since, f(x) is continuous at x=0 limx→0f(x)=f(0) limx→01−cos2x+sin2x√x2+1−1 =limx→02sin2xx2(√x2+1+1) =2limx→0(sinxx)2limx→0(√x2+1+1) =2×1×2=4 ∴f(0)=4