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Question Number 68171 by naka3546 last updated on 06/Sep/19

Commented by mathmax by abdo last updated on 06/Sep/19

$$$$$$\sqrt{5-2\sqrt{x}}isnotdefinedon+\mathrm{\infty }!$$

Commented by mathmax by abdo last updated on 07/Sep/19

$$ithinktheQhereisfind{lim}_{x\to +\mathrm{\infty }}\frac{\sqrt{2\sqrt{x}+5}-\sqrt{2\sqrt{x}-5}}{\sqrt{x}-1}={lim}_{x\to +\mathrm{\infty }}f\left(x\right)$$$$lettreatthisweusetbechangement\sqrt{x}=t$$$${lim}_{x\to +\mathrm{\infty }}f\left(x\right)={lim}_{t\to +\mathrm{\infty }}\frac{\sqrt{2t+5}-\sqrt{2t-5}}{t-1}$$$$={lim}_{t\to +\mathrm{\infty }}\frac{2t+5-2t+5}{(t-1)(\sqrt{2t+5}+\sqrt{2t-5})}={lim}_{t\to +\mathrm{\infty }}\frac{10}{(t-1)(\sqrt{2t+5}+\sqrt{2t-5)}})$$$$=0$$$$$$

Answered by Kunal12588 last updated on 06/Sep/19

$$\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{5+2\sqrt{x}}-\sqrt{5-2\sqrt{x}}}{\sqrt{x}-1}$$$$\sqrt{x}=t;x\to \mathrm{\infty }\Rightarrow t\to \mathrm{\infty }$$$$\underset{t\to \mathrm{\infty }}{lim}\frac{\sqrt{5+2t}-\sqrt{5-2t}}{t-1}$$$$=\underset{t\to \mathrm{\infty }}{lim}\frac{\sqrt{\frac{5}{t^2}+\frac{2}{t}}-\sqrt{\frac{5}{t^2}-\frac{2}{t}}}{1-\frac{1}{t}}$$$$=\frac{0}{1}=0$$

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