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Question Number 136258 by EDWIN88 last updated on 20/Mar/21

$$\underset{x\to 2\mathrm{a}}{\lim }\frac{\sqrt{\mathrm{x}-2\mathrm{a}}+\sqrt{\mathrm{x}}-\sqrt{2\mathrm{a}}}{\sqrt{{\mathrm{x}}^2-{4\mathrm{a}}^2}}=?\mathrm{where}\mathrm{a}>0$$

Answered by liberty last updated on 20/Mar/21

$$L^{\prime }H\ddot{opital}$$ $$=\underset{x\to 2a}{\lim }\frac{1}{\sqrt{x+2a}}.\underset{x\to 2a}{\lim }\frac{\sqrt{x-2a}+\sqrt{x}-\sqrt{2a}}{\sqrt{x-2a}}$$ $$=\frac{1}{2\sqrt{a}}.\underset{x\to 2a}{\lim }\frac{\frac{1}{2\sqrt{x-2a}}+\frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x-2a}}}$$ $$=\frac{1}{2\sqrt{a}}.\underset{x\to 2a}{\lim }\left(\frac{\sqrt{x}+\sqrt{x-2a}}{\sqrt{x}\sqrt{x-2a}}\right).\sqrt{x-2a}$$ $$=\frac{1}{2\sqrt{a}}.\underset{x\to 2a}{\lim }\left(\frac{\sqrt{x}+\sqrt{x-2a}}{\sqrt{x}}\right)$$ $$=\frac{1}{2\sqrt{a}}.\frac{\sqrt{2a}+0}{\sqrt{2a}}=\frac{1}{2\sqrt{a}}$$

Answered by EDWIN88 last updated on 20/Mar/21

$$\mathrm{Without}{\mathrm{L}}^{\prime }\mathrm{H}\ddot{\mathrm{o}pital}$$ $$\underset{x\to 2\mathrm{a}}{\lim }\frac{1}{\sqrt{\mathrm{x}+2\mathrm{a}}}.\underset{x\to 2\mathrm{a}}{\lim }\frac{(\sqrt{\mathrm{x}-2\mathrm{a}}+(\sqrt{\mathrm{x}}-\sqrt{2\mathrm{a}})}{\sqrt{\mathrm{x}-2\mathrm{a}}}$$ $$=\frac{1}{2\sqrt{\mathrm{a}}}.\underset{x\to 2\mathrm{a}}{\lim }(1+\frac{\sqrt{\mathrm{x}}-\sqrt{2\mathrm{a}}}{\sqrt{\mathrm{x}-2\mathrm{a}}})$$ $$=\frac{1}{2\sqrt{\mathrm{a}}}.(1+\underset{x\to 2\mathrm{a}}{\lim }\frac{\mathrm{x}-2\mathrm{a}}{(\sqrt{\mathrm{x}}+\sqrt{2\mathrm{a}})\sqrt{\mathrm{x}-2\mathrm{a}}})$$ $$=\frac{1}{2\sqrt{\mathrm{a}}}.(1+\underset{x\to 2\mathrm{a}}{\lim }\frac{\sqrt{\mathrm{x}-2\mathrm{a}}}{\sqrt{\mathrm{x}}+\sqrt{2\mathrm{a}}})=\frac{1}{2\sqrt{\mathrm{a}}}.(1+0)$$ $$=\frac{1}{2\sqrt{\mathrm{a}}}.$$

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