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Question Number 110988 by Rio Michael last updated on 01/Sep/20

$$\mathrm{Evaluate}\mathrm{without}\mathrm{using}{\mathrm{L}}^{\prime }{\mathrm{hopital}}^{\prime }\mathrm{s}\mathrm{rule}$$$$\underset{x\to 4}{\lim }\frac{\sqrt{x}-2}{x-4}$$

Answered by Rasheed.Sindhi last updated on 01/Sep/20

$$\underset{x\to 4}{\lim }\frac{\sqrt{x}-2}{x-4}$$$$\frac{\sqrt{x}-2}{{\left(\sqrt{x}\right)}^2-{\left(2\right)}^2}=\frac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}$$$$=\frac{1}{(\sqrt{x}+2)}$$$$\underset{x\to 4}{\lim }\frac{\sqrt{x}-2}{x-4}=\underset{x\to 4}{\lim }\frac{1}{(\sqrt{x}+2)}=\frac{1}{\sqrt{4}+2}$$$$=\frac{1}{4}$$$$$$

Commented by Rio Michael last updated on 01/Sep/20

$$\mathrm{correct}\mathrm{sir}.$$

Answered by malwan last updated on 02/Sep/20

$$\underset{x\to 4}{lim}\frac{\sqrt{x}-2}{x-4}\times \frac{\sqrt{x}+2}{\sqrt{x}+2}$$$$=\underset{x\to 4}{lim}\frac{x-4}{(x-4)(\sqrt{x}+2)}$$$$=\underset{x\to 4}{lim}\frac{1}{\sqrt{x}+2}=\frac{1}{\sqrt{4}+2}=\frac{1}{4}$$

Answered by bemath last updated on 01/Sep/20

$$\mathrm{let}\sqrt{\mathrm{x}}=\mathrm{b}\Rightarrow \mathrm{x}={\mathrm{b}}^2$$$$\underset{\mathrm{b}\to 2}{\lim }\frac{\mathrm{b}-2}{{\mathrm{b}}^2-4}=\underset{\mathrm{b}\to 2}{\lim }\frac{1}{\mathrm{b}+2}=\frac{1}{4}$$

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