Question

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

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

Hint:

First simplify the expression inside the limit completely, then evaluate the limit, and finally differentiate the resulting function.

Answer 1

The Correct Option is C

Step 1: Simplify the expression inside the limit.
We have
\[ \lim_{y\to 2}\frac{1}{y-2}\left(\frac{1}{x}-\frac{1}{x+y-2}\right) \] First simplify the bracket:
\[ \frac{1}{x}-\frac{1}{x+y-2} = \frac{x+y-2-x}{x(x+y-2)} \] \[ = \frac{y-2}{x(x+y-2)} \]

Step 2: Substitute this into the limit.
\[ \lim_{y\to 2}\frac{1}{y-2}\cdot \frac{y-2}{x(x+y-2)} \] Cancel \(y-2\):
\[ = \lim_{y\to 2}\frac{1}{x(x+y-2)} \]

Step 3: Evaluate the limit.
Now put \(y=2\):
\[ \frac{1}{x(x+2-2)} = \frac{1}{x^2} \]

Step 4: Differentiate with respect to \(x\).
Now,
\[ \frac{d}{dx}\left(\frac{1}{x^2}\right) = \frac{d}{dx}\left(x^{-2}\right) \] \[ =-2x^{-3} \] \[ =-\frac{2}{x^3} \]

Step 5: Final conclusion.
Hence,
\[ \boxed{-\frac{2}{x^3}} \]