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If $\mathrm{f}(\mathrm{x})=\sqrt{25-\mathrm{x}^{2}}$, then what is $\operatorname{Lim}_{\mathrm{x} \rightarrow 1} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(1)}{\mathrm{x}-1}$ equal to ?
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Verified Answer
The correct answer is:
$-\frac{1}{\sqrt{24}}$
$\mathrm{f}(\mathrm{x})=\sqrt{25-\mathrm{x}^{2}}$
$\mathrm{f}(1)=\sqrt{24}$
$\Rightarrow \lim _{\mathrm{x} \rightarrow 1} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(1)}{\mathrm{x}-1}$
$\therefore$ It is $\frac{0}{0}($ undefined condition $)$ so using L'hospital's
rule
$\Rightarrow \lim _{x \rightarrow 1} \frac{\mathrm{f}^{\prime}(\mathrm{x})-0}{1}=\lim _{\mathrm{x} \rightarrow 1}\left(\sqrt{25-\mathrm{x}^{2}}\right)^{\prime}$
$\Rightarrow \lim _{x \rightarrow 1} \frac{1}{2} \times \frac{1}{\sqrt{25-x^{2}}}(-2 x)$
$=\frac{1}{2} \times \frac{1}{\sqrt{25-(1)^{2}}} \times(-2)$
$=-\frac{1}{\sqrt{24}}$
$\mathrm{f}(1)=\sqrt{24}$
$\Rightarrow \lim _{\mathrm{x} \rightarrow 1} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(1)}{\mathrm{x}-1}$
$\therefore$ It is $\frac{0}{0}($ undefined condition $)$ so using L'hospital's
rule
$\Rightarrow \lim _{x \rightarrow 1} \frac{\mathrm{f}^{\prime}(\mathrm{x})-0}{1}=\lim _{\mathrm{x} \rightarrow 1}\left(\sqrt{25-\mathrm{x}^{2}}\right)^{\prime}$
$\Rightarrow \lim _{x \rightarrow 1} \frac{1}{2} \times \frac{1}{\sqrt{25-x^{2}}}(-2 x)$
$=\frac{1}{2} \times \frac{1}{\sqrt{25-(1)^{2}}} \times(-2)$
$=-\frac{1}{\sqrt{24}}$
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