Question ID: #651
Let $f:R\rightarrow(0,\infty)$ be a twice differentiable function such that $f(3)=18,$ $f'(3)=0$ and $f”(3)=4$. Then
$$ \lim_{x \to 1} \left( log_{e} \left( \frac{f(2+x)}{f(3)} \right)^{\frac{18}{(x-1)^{2}}} \right) $$
is equal to:
- (1) 1
- (2) 9
- (3) 2
- (4) 18
Solution:
$$L = \lim_{x \to 1} \log_e \left[ \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-1)^2}} \right]$$
$$\therefore L = \lim_{x \to 1} \frac{18}{(x-1)^2} \ln \left( \frac{f(x+2)}{18} \right)$$
$$L = 18 \lim_{x \to 1} \frac{\ln(f(x+2)) – \ln(18)}{(x-1)^2}$$
Apply L’Hopital’s Rule :
$$\therefore L = 18 \lim_{x \to 1} \frac{\frac{1}{f(x+2)} \cdot f'(x+2)}{2(x-1)}$$
$$\therefore L = 9 \lim_{x \to 1} \frac{f'(x+2)}{f(x+2)(x-1)}$$+
$$\therefore L = 9 \lim_{x \to 1} \frac{f'(x+2)}{f(3)(x-1)}$$
$$\therefore L = 9 \lim_{x \to 1} \frac{f'(x+2)}{18(x-1)}$$
$$\therefore L = \frac{9}{18} \lim_{x \to 1} \frac{f'(x+2)}{(x-1)}$$
Apply L’Hopital’s Rule again.
$$\therefore L = \frac{1}{2} \lim_{x \to 1} \frac{f”(x+2)}{1}$$
Substitute $x=1$:
$$\therefore L = \frac{1}{2} \left[ \frac{f”(3)}{1} \right]$$
$$\therefore L = \frac{1}{2} [ 4]$$
$$\therefore L = 2$$
Ans. (3)
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