Let f: R to (0, infty)$f: R \to (0, \infty)$ be a twice differentiable function such that f(3) = 18$f(3) = 18$ , f'(3) = 0$f'(3) = 0$ and f''(3) = 4$f''(3) = 4$ . Then lim_x to 1 left( log_e left( fracf(2 + x)f(3) right)^frac18(x-1)^2 right)$\lim_{x \to 1} \left( \log_{e} \left( \frac{f(2 + x)}{f(3)} \right)^{\frac{18}{(x-1)^2}} \right)$ is equal to:

### Related Formula For a limit of 1^infty$1^{\infty}$ form, lim_x to a [g(x)]^h(x)$\lim_{x \to a} [g(x)]^{h(x)}$ equals: e^lim_x to a h(x)[g(x) - 1]$$e^{\lim_{x \to a} h(x)[g(x) - 1]}$$ ### Core Logic Let T = lim_x to 1 left( fracf(x + 2)f(3) right)^frac18(x - 1)^2$T = \lim_{x \to 1} \left( \frac{f(x + 2)}{f(3)} \right)^{\frac{18}{(x - 1)^2}}$. As x to 1$x \to 1$, fracf(x+2)f(3) to fracf(3)f(3) = 1$\frac{f(x+2)}{f(3)} \to \frac{f(3)}{f(3)} = 1$. The exponent goes to infty$\infty$. This is a standard 1^infty$1^{\infty}$ form. T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)f(3) right)$$T = e^{\lim_{x \to 1} \frac{18}{(x - 1)^2} \left( \frac{f(x + 2) - f(3)}{f(3)} \right)}$$ ### Step 1: Simplify Exponent Limit Given f(3) = 18$f(3) = 18$: T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)18 right)$$T = e^{\lim_{x \to 1} \frac{18}{(x - 1)^2} \left( \frac{f(x + 2) - f(3)}{18} \right)}$$ T = e^lim_x to 1 fracf(x + 2) - f(3)(x - 1)^2$$T = e^{\lim_{x \to 1} \frac{f(x + 2) - f(3)}{(x - 1)^2}}$$ This is a frac00$\frac{0}{0}$ form limit. ### Step 2: Apply L'Hopital's Rule Differentiate numerator and denominator w.r.t x$x$: T = e^lim_x to 1 fracf'(x + 2)2(x - 1)$$T = e^{\lim_{x \to 1} \frac{f'(x + 2)}{2(x - 1)}}$$ This is still a frac00$\frac{0}{0}$ form since f'(3) = 0$f'(3) = 0$. Apply L'Hopital's Rule again: T = e^lim_x to 1 fracf''(x + 2)2$$T = e^{\lim_{x \to 1} \frac{f''(x + 2)}{2}}$$ Substitute x = 1$x = 1$: T = e^fracf''(3)2$$T = e^{\frac{f''(3)}{2}}$$ ### Step 3: Final Calculation Given f''(3) = 4$f''(3) = 4$: T = e^frac42 = e^2$$T = e^{\frac{4}{2}} = e^2$$ The question asks for log_e(T)$\log_e(T)$: log_e(T) = log_e(e^2) = 2$$\log_e(T) = \log_e(e^2) = 2$$ ### Pattern Recognition When expanding f(x)$f(x)$ around an extrema (f'(a)=0$f'(a)=0$) inside a 1^infty$1^{\infty}$ limit form, double L'Hopital isolates the second derivative divided by the Taylor factorial (2! = 2$2! = 2$). The limit elegantly shrinks directly to f''(a)/2$f''(a)/2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives Class 12 Maths: Limits, Continuity and Differentiability