1. Tardigrade
  2. Question
  3. Mathematics
  4. lim h → 0 ( f (2 h + 2 + h2 ) - f (2)/ f ( h - h2 + 1) - f (1)) , given thta f ' (2) = 6 and f ' (1) = 4.

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. $ lim_{ h \to 0 } \frac{ f (2 h + 2 + h^2 ) - f \, (2)}{ f \, ( h - h^2 + 1) - f (1)} $, given thta f ' (2) = 6 and f ' (1) = 4.

IIT JEEIIT JEE 2003

Solution:

Here, $ lim_{ h \to 0 } \frac{ f (2 h + 2 + h^2 ) - f \, (2)}{ f \, ( h - h^2 + 1) - f (1)} $ $$ [ $ \because $ f ' (2) = 6 and f ' (1) = 4, given ] Applying L'Hospital's rule, = $ lim_{ h \to 0 } \frac{ \{ f \, ' (2 h + 2 + h^2 ) \} . (2 + 2h) - 0 }{ \{ f \, ' ( h - h^2 + 1 ) \} . (1 - 2h) - 0 } = \frac{ f \, ' (2) . 2 }{ f \, ' \, (1) . 1 } $ = $ \frac{ 6 . 2}{ 4. 1} = 3 $ $$ [using f ' (2) = 6 and f ' (1) = 4]