Let $$f: R \rightarrow (0, \infty)$$ be a twice differentiable function such that f(3) = 18, f'(3) = 0 and f" (3) = 4. Then $$\lim_{x \rightarrow 1}\left(\log_{a}\left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^{2}}}\right)$$ ls equal to :

Let the limit be L. We are given:

$$\lim_{x \to 1} \left( \log_e \left( \frac{f(2+x)}{f(3)} \right)^{\frac{18}{(x-1)^2}} \right)$$

Using the property $$\log(a^b)=b\log(a)$$, we can bring the exponent down:

L = $$\lim_{x\to1}\frac{18}{(x-1)^2}\ln\left(\frac{f(2+x)}{f(3)}\right)$$ L = 18 $$\lim_{x\to1}\frac{\ln(f(2+x))-\ln(f(3))}{(x-1)^2}$$

The denominator $$(x-1)^2 \to 0$$.

The numerator $$\ln(f(2+1)) - \ln(f(3)) = \ln(f(3)) - \ln(f(3)) = 0$$.

Since this is a $$\frac{0}{0}$$ indeterminate form, we can apply L'Hôpital's Rule.

Differentiate the numerator and denominator with respect to x:

$$\frac{d}{dx} [\ln(f(2+x)) - \ln(f(3))] = \frac{f'(2+x)}{f(2+x)}$$ $$\frac{d}{dx} [(x-1)^2]$$= 2(x-1)

So the limit becomes:

L = 18 $$\lim_{x \to 1} \frac{f'(2+x)}{2(x-1) f(2+x)}$$ L = 9 $$\lim_{x \to 1} \frac{f'(2+x)}{(x-1) f(2+x)}$$

At x=1, the expression is still $$\frac{0}{0}$$because f'(3) = 0. Differentiating again:

• Numerator: $$\frac{d}{dx} f'(2+x)$$ = $$f''(2+x)$$
• Denominator: Using the product rule on $$(x-1)f(2+x)$$:$$\frac{d}{dx} [(x-1)f(2+x)] = 1 \cdot f(2+x) + (x-1)f'(2+x)$$

Now substitute x = 1 into the new expression:

L = 9 $$\cdot \frac{f''(3)}{f(3) + (1-1)f'(3)}$$ L = 9 $$\cdot \frac{f''(3)}{f(3)}$$

We are given:

• f(3) = 18
• f''(3) = 4

L = 9 $$\cdot \left( \frac{4}{18} \right)$$ L = 9$$\cdot \frac{2}{9}$$L = 2