This limit is 1∞ form
∴x→0lim(1+f(2−x)−f(2)1+f(3+x)−f(3))x1=ex→0limx1(1+f(2−x)−f(2)1+f(3+x)−f(3)−1)
=ex→0limx1(1+f(2−x)−f(2)f(3+x)−f(3)−f(2−x)+f(2))(00form)
=ex→0limx(−f′(2−x))+(1+f(2−x)−f(2))f′(3+x)+f′(2−x)
=e0=1.
Let f:R→R be a differentiable function satisfying f′(3)+f′(2)=0. Then x→0lim(1+f(2−x)−f(2)1+f(3+x)−f(3))x1 is equal to
Held on 8 Apr 2019 · Verified 6 Jul 2026.
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