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Question Number 39078 by Cheyboy last updated on 02/Jul/18

$$\mathit{W}\mathit{i}\mathit{t}\mathit{h}\mathit{o}\mathit{u}\mathit{t}\mathit{u}\mathit{s}\mathit{i}\mathit{n}\mathit{g}{\mathit{l}}^{\prime }\mathit{h}\mathit{o}\mathit{p}\mathit{i}\mathit{t}\mathit{a}\mathit{l}$$$$\mathit{f}\mathit{i}\mathit{n}\mathit{d}\underset{x\to 3}{\lim }\frac{\sqrt{9-{\mathit{x}}^2}}{\mathit{x}-3}$$$$$$

Commented by math khazana by abdo last updated on 03/Jul/18

$${lim}_{x\to 3^{-}}\frac{\sqrt{9-x^2}}{x-3}={lim}_{x\to 3^{-}}-\frac{\sqrt{9-x^2}}{3-x}$$$$={lim}_{x\to 3^{-}}-\sqrt{\frac{9-x^2}{{(3-x)}^2}}={lim}_{x\to 3^{-}}-\sqrt{\frac{3+x}{3-x}}$$$$=-\mathrm{\infty }.$$

Commented by orlandorap123 last updated on 03/Aug/18

$$$$$$\underset{x\to 3}{\lim }\frac{\sqrt{9-{\mathit{x}}^2}}{\mathit{x}-3}$$$$\underset{x\to 3}{\lim }\frac{(3-x)(3+x)}{-(x-3)\sqrt{9-x^2}}$$$$\underset{x\to 3}{\lim }\frac{(3+x)}{-\sqrt{9-x^2}}$$$$\mathrm{evaluando}$$$$\underset{x\to 3}{\lim }\frac{\left(6\right)}{-0}=-\mathrm{\infty }$$$$\mathit{f}\mathit{i}\mathit{n}\mathit{d}\underset{x\to 3}{\lim }\frac{\sqrt{9-{\mathit{x}}^2}}{\mathit{x}-3}=-\mathrm{\infty }$$

Answered by ajfour last updated on 02/Jul/18

$$lefthandlimit=\underset{h\to 0}{\lim }\frac{\sqrt{9-{(3-h)}^2}}{(3-h)-3}$$$$=\underset{h\to 0}{\lim }\frac{\sqrt{6h-h^2}}{-h}=-\mathrm{\infty }$$

Commented by Cheyboy last updated on 02/Jul/18

$$Thankyousir$$

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