Let $f: R \rightarrow R$ be a differentiable function satisfying $f^{\prime}(3)+f^{\prime}(2)=0$. Then $\lim_{x \rightarrow \infty}\left(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\right)^{\frac{1}{x}}$ is equal to:

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Limits

Let

f : R \to R

be a differentiable function satisfying

f^{'} (3) + f^{'} (2) = 0

. Then

lim_{x \to \infty} (\frac{1 + f ( 3 + x ) - f ( 3 )}{1 + f ( 2 - x ) - f ( 2 )})^{\frac{1}{x}}

is equal to:

Answer: (B)

This is a

1^{\infty}

form, handled by

lim [g]^{h} = e^{l i m (g - 1) h}

. Here

g - 1 = \frac{f ( 3 + x ) - f ( 3 ) - [ f ( 2 - x ) - f ( 2 ) ]}{1 + f ( 2 - x ) - f ( 2 )}

and

h = \frac{1}{x}

, so the exponent is

lim_{x \to \infty} \frac{f ( 3 + x ) - f ( 3 ) - f ( 2 - x ) + f ( 2 )}{x [ 1 + f ( 2 - x ) - f ( 2 ) ]}

. For large

x

the growth of

f (3 + x) - f (3)

is governed by

f^{'} (3)

and that of

f (2 - x) - f (2)

- f^{'} (2)

; applying L'Hopital and using the given

f^{'} (3) + f^{'} (2) = 0

makes the exponent vanish. So the limit is

e^{0} = 1

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option