If G(x) = - √ 25 − x 2 then lim x → 1 G ( x ) − G ( 1 ) x − 1 has the value Options: A . &nbsp 1 24 B . &nbsp 1 5 C . &nbsp − √ 24 D . &nbsp 1 2 √ 6

Answer: Option D : D lim x → 1 − √ 25 − x 2 − ( − √ 24 ) x − 1 = lim x → 1 √ 24 − √ 25 − x 2 x − 1 × √ 24 + √ 25 − x 2 √ 24 + √ 25 − x 2 lim x → 1 x 2 − 1 ( x − 1 ) [ √ 24 + √ 25 − x 2 ] = 2 2 √ 24 = 1 2 √ 6

If G(x) = - √25−x2 then limx→1G(x)−G(1)x−1

Discussion Forum : Limits Continuity And Differentiability

Question -

If G(x) = - $\sqrt{25 - x^{2}}$ then $lim x \to 1 \frac{G (x) - G (1)}{x - 1}$ has the value

Options:

A .

\frac{1}{24}

B .

\frac{1}{5}

C .

- \sqrt{24}

D .

\frac{1}{2 \sqrt{6}}

Answer: Option D
:
D

lim x \to 1 \frac{- \sqrt{25 - x^{2}} - (- \sqrt{24})}{x - 1}

= lim x \to 1 \frac{\sqrt{24} - \sqrt{25 - x^{2}}}{x - 1} \times \frac{\sqrt{24} + \sqrt{25 - x^{2}}}{\sqrt{24} + \sqrt{25 - x^{2}}}

lim x \to 1 \frac{x^{2} - 1}{(x - 1) [\sqrt{24} + \sqrt{25 - x^{2}}]} = \frac{2}{2 \sqrt{24}} = \frac{1}{2 \sqrt{6}}

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Discussion Forum : Limits Continuity And Differentiability

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