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 $[\cdot]$ denote the greatest integer function, and let $f(x) = \min \{\sqrt{2}x, x^2\}$.
Let $S = \{x \in (-2, 2) : \text{the function } g(x) = |x|[x^2] \text{ is discontinuous at } x\}$.
Then $\sum_{x \in S} f(x)$ equals - If the function $f(x) = \frac{e^x\left(e^{\tan x - x} - 1\right) + \log_e(\sec x + \tan x) - x}{\tan x - x}$ is continuous at $x = 0$, then the value of $f(0)$ is equal to
- 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 ______
- Consider the following three statements for the function $f : (0, \infty) \rightarrow \mathbb{R}$ defined by $f(x) = |\log_e x| - |x - 1|$:
(I) $f$ is differentiable at all $x > 0$.
(II) $f$ is increasing in $(0, 1)$.
(III) $f$ is decreasing in $(1, \infty)$.
Then. - If the domain of the function $f(x) = \sin^{-1} \left( \frac{1}{x^2 - 2x - 2} \right)$ is $(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$, then $\alpha + \beta + \gamma + \delta$ is equal to
Let $[t]$ denote the greatest integer less than or equal to $t$. If the function
$f(x) = \begin{cases} b^2 \sin \left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\cos x + \sin x) \cos x \right] \right), & x < 0 \\ \frac{\sin x - \frac{1}{2} \sin 2x}{x^3}, & x > 0 \\ a, & x = 0 \end{cases}$
is continuous at $x = 0$, then $a^2 + b^2$ is equal to
- 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 :
- 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 ____________.
- 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, - 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 :
Important Questions from Differential Calculus
- Let $[\cdot]$ denote the greatest integer function, and let $f(x) = \min \{\sqrt{2}x, x^2\}$.
Let $S = \{x \in (-2, 2) : \text{the function } g(x) = |x|[x^2] \text{ is discontinuous at } x\}$.
Then $\sum_{x \in S} f(x)$ equals - If the function $f(x) = \frac{e^x\left(e^{\tan x - x} - 1\right) + \log_e(\sec x + \tan x) - x}{\tan x - x}$ is continuous at $x = 0$, then the value of $f(0)$ is equal to
- 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 ______
- Consider the following three statements for the function $f : (0, \infty) \rightarrow \mathbb{R}$ defined by $f(x) = |\log_e x| - |x - 1|$:
(I) $f$ is differentiable at all $x > 0$.
(II) $f$ is increasing in $(0, 1)$.
(III) $f$ is decreasing in $(1, \infty)$.
Then. - If the domain of the function $f(x) = \sin^{-1} \left( \frac{1}{x^2 - 2x - 2} \right)$ is $(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$, then $\alpha + \beta + \gamma + \delta$ is equal to
