Let f : R to R be a differentiable function s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The expression is of the form $1^\infty$ as $x 	o 0$.Let $L = \mathop {\lim }\limits_{x 	o 0} {\left( {\frac{{1 + f\left( {3 + x} ight) - f\le

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The expression is of the form $1^\infty$ as $x 	o 0$.Let $L = \mathop {\lim }\limits_{x 	o 0} {\left( {\frac{{1 + f\left( {3 + x} ight) - f\left( 3 ight)}}{{1 + f\left( {2 - x} ight) - f\left( 2 ight)}}} ight)^{\frac{1}{x}}}$.Using the formula $\mathop {\lim }\limits_{x 	o a} {\left( {f(x)} ight)^{g(x)}} = e^{\mathop {\lim }\limits_{x 	o a} g(x)(f(x) - 1)}$,we get:$L = e^{\mathop {\lim }\limits_{x 	o 0} \frac{1}{x} \left( \frac{1 + f(3+x) - f(3)}{1 + f(2-x) - f(2)} - 1 ight)}$$L = e^{\mathop {\lim }\limits_{x 	o 0} \frac{1}{x} \left( \frac{f(3+x) - f(3) - f(2-x) + f(2)}{1 + f(2-x) - f(2)} ight)}$Applying $L$'Hopital's rule to the exponent:Exponent $= \mathop {\lim }\limits_{x 	o 0} \frac{f'(3+x) + f'(2-x)}{1 + f(2-x) - f(2) - xf'(2-x)} = \frac{f'(3) + f'(2)}{1} = 0$.Thus,$L = e^0 = 1$.

Let $f : R \to R$ be a differentiable function satisfying $f''(3) + f'(2) = 0$. Then $\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + f\left( {3 + x} \right) - f\left( 3 \right)}}{{1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right)^{\frac{1}{x}}}$ is equal to

Similar Questions

Suppose $a$ is a positive real number such that $a^5-a^3+a=2$. Then,

Consider the function $f(x) = x \cos x - \sin x$. Identify the correct statement.

Let $f:[0, \infty) \rightarrow [0, 3]$ be a function defined by
$f(x) = \begin{cases} \max \{\sin t : 0 \leq t \leq x\}, & 0 \leq x \leq \pi \\ 2 + \cos x, & x > \pi \end{cases}$
Then which of the following is true?

The function $f(x) = \begin{cases} |x - 3| & x \geqslant 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4} & x < 1 \end{cases}$ is :

If $f:R \to R$ and $f(x)$ is a polynomial function of degree $10$ such that $f(x)=0$ has all real and distinct roots,then the equation $(f'(x))^2 - f(x)f''(x) = 0$ has:

Vedclass Test Series

Exam Paper Generator

Online Exam Module