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Question Number 131388 by Eric002 last updated on 04/Feb/21

$$\underset{x\to 4}{lim}\frac{\sqrt{2x^2-16}+4}{4x^2-16x}-\frac{1}{\sqrt{2x^2-16}-4}$$

Answered by bemath last updated on 04/Feb/21

$$\underset{x\to 4}{\lim }(\frac{\sqrt{2x^2-16}+4}{4x^2-16x}-\frac{1}{\sqrt{2x^2-16}-4})=?$$$$(\bullet )\frac{1}{\sqrt{2x^2-16}-4}=\frac{\sqrt{2x^2-16}+4}{2x^2-32}$$$$(\bullet \bullet )\underset{x\to 4}{\lim }\frac{\sqrt{2x^2-16}+4}{4x^2-16x}-\frac{\sqrt{2x^2-16}+4}{2x^2-32}$$$$\underset{x\to 4}{\lim }(\sqrt{2x^2-16}+4)\times \underset{x\to 4}{\lim }(\frac{1}{4x^2-16x}-\frac{1}{2x^2-32})$$$$=8\times \underset{x\to 4}{\lim }(\frac{1}{4x(x-4)}-\frac{1}{2(x-4)(x+4)})$$$$=8\times \underset{x\to 4}{\lim }(\frac{x+4}{4x(x-4)(x+4)}-\frac{2x}{4x(x-4)(x+4)})$$$$=8\times \underset{x\to 4}{\lim }\left(\frac{-(x-4)}{4x(x-4)(x+4)}\right)$$$$=8\times \underset{x\to 4}{\lim }\left(\frac{-1}{4x(x+4)}\right)=-\frac{8}{16\times 8}$$$$=-\frac{1}{16}$$$$$$

Commented by bramlexs22 last updated on 04/Feb/21

$$$$$$\mathrm{Wow}\mathrm{es}\mathrm{ist}\mathrm{so}\mathrm{cool}\mathrm{Sir}$$

Commented by Eric002 last updated on 04/Feb/21

$$welldonesir$$

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