frac{d}{dx} left( lim_{y to 2} frac{1}{y-2} l | vedclass

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

Question

(C) First,simplify the expression inside the limit: $\frac{1}{y-2} \left( \frac{1}{x} - \frac{1}{x+y-2} \right) = \frac{1}{y-2} \left( \frac{(x+y-2) -

Vedclass · Accepted Answer

$\frac{-2}{x^3}$

Vedclass · Answer

(C) First,simplify the expression inside the limit: $\frac{1}{y-2} \left( \frac{1}{x} - \frac{1}{x+y-2} ight) = \frac{1}{y-2} \left( \frac{(x+y-2) - x}{x(x+y-2)} ight) = \frac{1}{y-2} \left( \frac{y-2}{x(x+y-2)} ight) = \frac{1}{x(x+y-2)}$.Now,evaluate the limit as $y 	o 2$: $\lim_{y 	o 2} \frac{1}{x(x+y-2)} = \frac{1}{x(x+2-2)} = \frac{1}{x^2}$.Finally,differentiate with respect to $x$: $\frac{d}{dx} \left( \frac{1}{x^2} ight) = \frac{d}{dx} (x^{-2}) = -2x^{-3} = \frac{-2}{x^3}$.

$\frac{d}{dx} \left( \lim_{y \to 2} \frac{1}{y-2} \left( \frac{1}{x} - \frac{1}{x+y-2} \right) \right) = $

Similar Questions

If $f(1)=1$ and $f^{\prime}(1)=3$,then the derivative of $f(f(f(x)))+(f(x))^2$ at $x=1$ is

The derivative of $e^{ax} \cos bx$ with respect to $x$ is $re^{ax} \cos(bx + \alpha)$,where $\alpha = \tan^{-1}(\frac{b}{a})$. When $a > 0, b > 0$,the value of $r$ is:

$[x]$ represents the greatest integer function. At $x = -1$,what is the value of $\frac{d}{dx} \sin(\pi[x])$?

If the derivative of $\tan^{-1}(a + bx)$ with respect to $x$ at $x = 0$ is $1$, then $a^6 - b^3 = $

If $f(x) = x^2 - 3x$,then the points at which $f(x) = f'(x)$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module