Let $f: \mathrm{R} \rightarrow(0, \infty)$ be a twice differentiable function such that $\mathrm{f}(3)=18, \mathrm{f}^{\prime}(3)=0$ and $\mathrm{f}^{\prime \prime}(3)=4$. Then $\lim _{x \rightarrow 1}\left(\log _{e}\left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^{2}}}\right)$ is equal to :

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Limits

Let

f : R \to (0, \infty)

be a twice differentiable function such that

f (3) = 18, f^{'} (3) = 0

and

f^{''} (3) = 4

.
Then

lim_{x \to 1} (lo g_{e} (\frac{f ( 2 + x )}{f ( 3 )})^{\frac{18}{( x - 1 ) ^{2}}})

is equal to :

Answer: (C)

The inner ratio tends to

1

, giving a

1^{\infty}

form whose logarithm is

lim_{x \to 1} \frac{18}{( x - 1 ) ^{2}} ln \frac{f ( x + 2 )}{18}

. Setting

h = x - 1

and expanding about

x = 3

f (3 + h) = 18 + \frac{f ^{''} ( 3 )}{2} h^{2} + \dots = 18 + 2 h^{2} + \dots

since

f^{'} (3) = 0

, so

ln \frac{f ( 3 + h )}{18} \approx \frac{h ^{2}}{9}

. Therefore the limit is

\frac{18}{h ^{2}} \cdot \frac{h ^{2}}{9} = 2

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option

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