If f(x) satisfies 97 f(x) + m fleft(frac{1}{x | vedclass

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

Question

(B) Given $f(x) = \lim_{n \rightarrow \infty} n(x^{1/n} - 1)$. Let $h = \frac{1}{n}$. As $n \rightarrow \infty$,$h \rightarrow 0$. $f(x) = \lim_{h \ri

Vedclass · Accepted Answer

$97$

Vedclass · Answer

(B) Given $f(x) = \lim_{n ightarrow \infty} n(x^{1/n} - 1)$. Let $h = \frac{1}{n}$. As $n ightarrow \infty$,$h ightarrow 0$. $f(x) = \lim_{h ightarrow 0} \frac{x^h - 1}{h}$. Using the standard limit formula $\lim_{h ightarrow 0} \frac{a^h - 1}{h} = \ln a$,we get $f(x) = \ln x$. Now,substitute $f(x) = \ln x$ into the given equation: $97 f(x) + m f\left(\frac{1}{x}ight) = 0$ $97 \ln x + m \ln\left(\frac{1}{x}ight) = 0$ Since $\ln\left(\frac{1}{x}ight) = -\ln x$,we have: $97 \ln x - m \ln x = 0$ $(97 - m) \ln x = 0$ For this to hold for all $x > 0$ (where $\ln x 
eq 0$),we must have $97 - m = 0$,which implies $m = 97$.

If $f(x)$ satisfies $97 f(x) + m f\left(\frac{1}{x}\right) = 0$,where $f(x) = \lim_{n \rightarrow \infty} n(x^{1/n} - 1)$ for $x > 0$,then the value of $m$ is:

Similar Questions

If $\lim _{x \rightarrow \infty}\left(\frac{11 x^3-3 x+4}{13 x^3-5 x^2-7}\right)=\frac{a}{b}$,then the value of $a+b$ equals:

$\lim _{x \rightarrow 0} \frac{\cos 4 x-4 \cos 2 x+3}{x^4} = $

Let $f(x) = \lim_{y \to 0} \frac{(1 - \cos(xy))\tan(xy)}{y^3}$. Then the number of solutions of the equation $f(x) = \sin x, x \in R$ is:

$\lim _{x \rightarrow \infty}\left(1+\frac{2}{x}\right)^{3 x}=$

Evaluate the given limit: $\mathop {\lim }\limits_{z \to 1} \frac{z^{1/3}-1}{z^{1/6}-1}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module