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Mathematics Question on Limits

limh→0f(2h+2+h2)−f (2)f (h−h2+1)−f(1)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.

A

does not exist

B

is equal to -3/2

C

is equal to 3/2

D

is equal to 3

Answer

does not exist

Explanation

Solution

Here, limh→0f(2h+2+h2)−f (2)f (h−h2+1)−f(1)lim_{ h \to 0 } \frac{ f (2 h + 2 + h^2 ) - f \, (2)}{ f \, ( h - h^2 + 1) - f (1)}
\hspace 25mm [ ∵\because f ' (2) = 6 and f ' (1) = 4, given ]
Applying L'Hospital's rule,
= limh→0f ′(2h+2+h2).(2+2h)−0f ′(h−h2+1).(1−2h)−0=f ′(2).2f ′ (1).1lim_{ 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 }
= 6.24.1=3\frac{ 6 . 2}{ 4. 1} = 3 \hspace15mm [using f ' (2) = 6 and f ' (1) = 4]