Evaluate the derivative lim┬(x→1)[1/(1-x) ∫_(x^2)^x▒(t^2-1)dt]
limx→1[11−x∫−x2x2(t2−1)dt]lim_{x\rightarrow1}[\dfrac{1}{1-x}\int_{-x^2}^{x^2}(t^2-1)dt]limx→1[1−x1∫−x2x2(t2−1)dt]
=limx→1[11−x(t33−t)∣−x2x2=lim_{x\rightarrow1}[\dfrac{1}{1-x}(\dfrac{t^3}{3}-t)|_{-x^2}^{x^2}=limx→1[1−x1(3t3−t)∣−x2x2
=limx→1[11−x(x63−x2+x63−x2)=lim_{x\rightarrow1}[\dfrac{1}{1-x}(\dfrac{x^6}{3}-x^2+\dfrac{x^6}{3}-x^2)=limx→1[1−x1(3x6−x2+3x6−x2)
=limx→1[11−x2x2(x33−1)=lim_{x\rightarrow1}[\dfrac{1}{1-x}2x^2(\dfrac{x^3}{3}-1)=limx→1[1−x12x2(3x3−1)
=2(1)2(13−1)1−1=∞=\dfrac{2(1)^2(\dfrac{1}{3}-1)}{1-1}=\infty=1−12(1)2(31−1)=∞
Hence The given limit does not exist.
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