WBJEE · Maths · Limits
\(f(x)\) is a differentiable function and given \(f^{\prime}(2)=6\) and \(f^{\prime}(1)=4\), then \(L=\lim _{h \rightarrow 0} \frac{f\left(2+2 h+h^2\right)-f(2)}{f\left(1+h-h^2\right)-f(1)}\)
- A does not exist
- B equal to -3
- C equal to 3
- D equal to \(3 / 2\)
Answer & Solution
Correct Answer
(C) equal to 3
Step-by-step Solution
Detailed explanation
Hint: \(\operatorname{It}_{h \rightarrow 0} \frac{f\left(2+2 h+h^2\right)-f(2)}{f\left(1+h-h^2\right)-f(1)} \quad[\div(\) form \()]\) by L. Hospital rule.…
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