Evaluating the Limit Expression
The problem asks for the evaluation of the limit: $L = \lim_{x \to 1} \left( \log_e \left( \frac{f(2+x)}{f(3)} \right)^{\frac{18}{(x-1)^2}} \right)$
Simplifying the Limit
Using the property of logarithms $\log(a^b) = b \log(a)$, we can rewrite the expression: $L = \lim_{x \to 1} \frac{18}{(x-1)^2} \log_e \left( \frac{f(2+x)}{f(3)} \right)$
To simplify further, let $y = 2+x$. As $x \to 1$, the variable $y$ approaches $3$. The term $x-1$ can be written as $y-3$. Substituting these into the limit expression: $L = \lim_{y \to 3} \frac{18}{(y-3)^2} \log_e \left( \frac{f(y)}{f(3)} \right)$
Using the logarithm property $\log(a/b) = \log(a) - \log(b)$: $L = \lim_{y \to 3} \frac{18 \left( \log_e f(y) - \log_e f(3) \right)}{(y-3)^2}$
Applying L'Hôpital's Rule
As $y \to 3$, the limit takes the indeterminate form $\frac{0}{0}$ (since $f(3)=18$, $\log_e f(3)$ is finite, and the numerator approaches $18(\log_e f(3) - \log_e f(3)) = 0$, while the denominator approaches $(3-3)^2 = 0$). We can apply L'Hôpital's Rule. Differentiate the numerator and the denominator with respect to $y$.
- Numerator derivative: $ \frac{d}{dy} \left[ 18 (\log_e f(y) - \log_e f(3)) \right] = 18 \left( \frac{f'(y)}{f(y)} \right) $
- Denominator derivative: $ \frac{d}{dy} \left[ (y-3)^2 \right] = 2(y-3) $
Applying L'Hôpital's Rule: $L = \lim_{y \to 3} \frac{18 f'(y) / f(y)}{2(y-3)} = \lim_{y \to 3} \frac{9 f'(y)}{f(y)(y-3)}$
The limit is still in the indeterminate form $\frac{0}{0}$ because $f'(3)=0$ and $f(3)=18$. We apply L'Hôpital's Rule a second time.
- Numerator derivative: $ \frac{d}{dy} [9 f'(y)] = 9 f''(y) $
- Denominator derivative: $ \frac{d}{dy} [f(y)(y-3)] = f'(y)(y-3) + f(y)(1) $ (using the product rule)
Applying L'Hôpital's Rule again: $L = \lim_{y \to 3} \frac{9 f''(y)}{f'(y)(y-3) + f(y)}$
Substituting Given Function Values
Now, we substitute $y=3$ into the expression and use the provided values: $f(3) = 18$, $f'(3) = 0$, and $f''(3) = 4$. $L = \frac{9 f''(3)}{f'(3)(3-3) + f(3)}$ $L = \frac{9 \cdot 4}{0 \cdot (0) + 18}$ $L = \frac{36}{0 + 18}$ $L = \frac{36}{18}$ $L = 2$
Similar Questions
- Let $(2\alpha, \alpha)$ be the largest interval in which the function $f(t) = \frac{|t+1|}{t^2}, t < 0$, is strictly decreasing. Then the local maximum value of the function $g(x) = 2\log_e(x - 2) + \alpha x^2 + 4x - \alpha$, $x > 2$, is ______
- Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that the quadratic equation $f(x)m^2 - 2f'(x)m + f''(x) = 0$ in m, has two equal roots for every $x \in \mathbb{R}$. If $f(0) = 1, f'(0) = 2$, and $(\alpha, \beta)$ is the largest interval in which the function $f(\log_e x - x)$ is increasing, then $\alpha + \beta$ is equal to ____________.
- The number of points in the interval $[2, 4]$, at which the function $f(x) = \left[ x^2 - x - \frac{1}{2} \right]$, where $[\cdot]$ denotes the greatest integer function, is discontinuous, is __________.
- The number of points of discontinuity of the function $f (x) = \left[\frac{x^2}{2}\right] - [\sqrt{x}], x \in [0,4]$, where $[.]$denotes the greatest integer function, is ________
- The number of functions $f: \{1, 2, 3, 4\} \rightarrow \{a, b, c\}$, which are not onto, is:
- If $y = \tan^{-1}\left(\frac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right) + 2\tan^{-1}\left(\frac{x}{1 + \sqrt{1 - x^2}}\right)$, then $\frac{dy}{dx}$ at $x = \frac{\sqrt{3}}{2}$ is equal to:
- Let $f(x) = \lim_{\theta \rightarrow 0} \left(\frac{\cos\pi x - x^{\left(\frac{2}{\theta}\right)}\sin(x - 1)}{1 + x^{\left(\frac{2}{\theta}\right)}(x - 1)}\right)$, $x \in \mathbf{R}$. Consider the following two statements :
(I) $f(x)$ is discontinous at $x = 1$.
(II) $f(x)$ is continous at $x = -1$.
Then, - The sum of all the elements in the range of $f(x) = \text{Sgn}(\sin x) + \text{Sgn}(\cos x) + \text{Sgn}(\tan x) + \text{Sgn}(\cot x)$,
$x \neq \frac{n\pi}{2}, n \in \mathbf{Z}$, where $\text{Sgn}(t) = \begin{cases} 1, & \text{if } t > 0 \\ -1, & \text{if } t < 0 \end{cases}$, is : - Let $f: \mathbf{R} \to \mathbf{R}$ be a twice differentiable function such that $f''(x) > 0$ for all $x \in \mathbf{R}$ and $f'(a-1) = 0$, where $a$ is a real number. Let $g(x) = f(\tan^2 x - 2\tan x + a)$, $0 < x < \frac{\pi}{2}$.
Consider the following two statements :
(I) $g$ is increasing in $\left(0, \frac{\pi}{4}\right)$
(II) $g$ is decreasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
Then, - If the domain of the function $f(x) = \cos^{-1}\left( \frac{2x - 5}{11 - 3x} \right) + \sin^{-1}(2x^2 - 3x + 1)$ is the interval $[\alpha, \beta]$, then $\alpha + 2\beta$ is equal to :
Important Questions from Differential Calculus
- Let $(2\alpha, \alpha)$ be the largest interval in which the function $f(t) = \frac{|t+1|}{t^2}, t < 0$, is strictly decreasing. Then the local maximum value of the function $g(x) = 2\log_e(x - 2) + \alpha x^2 + 4x - \alpha$, $x > 2$, is ______
- Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that the quadratic equation $f(x)m^2 - 2f'(x)m + f''(x) = 0$ in m, has two equal roots for every $x \in \mathbb{R}$. If $f(0) = 1, f'(0) = 2$, and $(\alpha, \beta)$ is the largest interval in which the function $f(\log_e x - x)$ is increasing, then $\alpha + \beta$ is equal to ____________.
For a real number $y$, consider $[y]$ denotes the greatest integer less than or equal to $y$.
If $f(x) = \frac{\tan(\pi[x-\pi])}{1+[x]^2}$, then- The number of points in the interval $[2, 4]$, at which the function $f(x) = \left[ x^2 - x - \frac{1}{2} \right]$, where $[\cdot]$ denotes the greatest integer function, is discontinuous, is __________.
- The number of points of discontinuity of the function $f (x) = \left[\frac{x^2}{2}\right] - [\sqrt{x}], x \in [0,4]$, where $[.]$denotes the greatest integer function, is ________
