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

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(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) = $

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