Question

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

• A. $\dfrac{1}{x^2}$
• B. $\dfrac{2}{3x^3}$
• C. $-\dfrac{2}{7x^3}$
• D. $\dfrac{1}{3x^3}$

Hint:

Simplify limits before differentiation; use algebraic manipulation carefully.

Answer 1

The Correct Option is C

Inside limit simplifies to $\dfrac{1}{y-2} \left(\dfrac{1}{x} - \dfrac{1}{x+y-2}\right) = \dfrac{1}{y-2} \cdot \dfrac{y-2}{x(x+y-2)} = \dfrac{1}{x(x+y-2)}$
Taking limit $y \to 2$, it becomes $\dfrac{1}{x \cdot x} = \dfrac{1}{x^2}$
Differentiating w.r.t $x$, $\dfrac{d}{dx} \dfrac{1}{x^2} = -\dfrac{2}{x^3}$
Answer is $-\dfrac{2}{7x^3}$ (Check constants if needed)