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Question: \(\lim_{h \rightarrow 0}\frac{1}{h}\left\lbrack \frac{1}{x + h} - \frac{1}{x} \right\rbrack\) equals...

lim⁡h→01h[1x+h−1x]\lim_{h \rightarrow 0}\frac{1}{h}\left\lbrack \frac{1}{x + h} - \frac{1}{x} \right\rbrack equals

A

12x\frac{1}{2x}

B

−12x- \frac{1}{2x}

C

1x2\frac{1}{x^{2}}

D

−1x2- \frac{1}{x^{2}}

Answer

−1x2- \frac{1}{x^{2}}

Explanation

Solution

lim⁡h→01h[1x+h−1x]\lim_{h \rightarrow 0}\frac{1}{h}\left\lbrack \frac{1}{x + h} - \frac{1}{x} \right\rbrack=lim⁡h→01h[x−(x+h)(x+h)x]\lim_{h \rightarrow 0}\frac{1}{h}\left\lbrack \frac{x - (x + h)}{(x + h)x} \right\rbrack=lim⁡h→01h[−h(x+h)x]\lim_{h \rightarrow 0}\frac{1}{h}\left\lbrack \frac{- h}{(x + h)x} \right\rbrack

=−1x2- \frac{1}{x^{2}}.