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

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. (A) does not exist (B) is equal to -3/2 (C) is equal to 3/2 (D) is equal to 3

Tardigrade Admin · Accepted Answer

Here, lim h → 0 ( f (2 h + 2 + h2 ) - f (2)/ f ( h - h2 + 1) - f (1)) [ ∵ f ' (2) = 6 and f ' (1) = 4, given ] Applying L'Hospital's rule, = lim h → 0 ( f ' (2 h + 2 + h2 ) . (2 + 2h) - 0 / f ' ( h - h2 + 1 ) . (1 - 2h) - 0 ) = ( f ' (2) . 2 / f ' (1) . 1 ) = ( 6 . 2/ 4. 1) = 3 [using f ' (2) = 6 and f ' (1) = 4]

Q. limh→0​f(h−h2+1)−f(1)f(2h+2+h2)−f(2)​, given thta f ' (2) = 6 and f ' (1) = 4.

does not exist

is equal to -3/2

is equal to 3/2

is equal to 3

Here, limh→0​f(h−h2+1)−f(1)f(2h+2+h2)−f(2)​ [ ∵ f ' (2) = 6 and f ' (1) = 4, given ] Applying L'Hospital's rule, = limh→0​{f′(h−h2+1)}.(1−2h)−0{f′(2h+2+h2)}.(2+2h)−0​=f′(1).1f′(2).2​ = 4.16.2​=3 [using f ' (2) = 6 and f ' (1) = 4]

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