f′(a)=2,f(a)=4
x→alimx−axf(a)−af(x)
⇒x→alim1f(a)−af′(x) (L'Hospitals rule)
=f(a)−af′(a)
=4−2a
JEE Main 2021 — Mathematics Calculus
Let f(x) be a differentiable function at x=a with f′(a)=2 and f(a)=4. Then x→alimx−axf(a)−af(x) equals:
Held on 26 Feb 2021 · Verified 6 Jul 2026.
a+4
2a−4
4−2a
2a+4
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