If f(1)=1, f′(1)=2, then write the value of limx→1√f(x)−1√x−1
Given
f(1)=1
f′(1)=2
Now,
limx→1√f(x)−1√x−1=limx→1f′(x)2√f(x)12√x
=limx→1f′(x)√x√f(x)=f′(1)√1√f(1)=2
If f(x)=−√25−x2, then find limx→1(f(x)−f(1)x−1)