lim-x-4x-4x-4x-4x-4x-

Question Number 117052 by bemath last updated on 09/Oct/20

\lim_{x \to \infty} \sqrt{4 x + \sqrt{4 x + \sqrt{4 x + \sqrt{4 x}}}} - \sqrt{4 x} = ?

Answered by bobhans last updated on 09/Oct/20

\lim_{x \to \infty} \sqrt{4 x (1 + \frac{\sqrt{4 x + \sqrt{4 x + \sqrt{4 x}}}}{4 x})} - \sqrt{4 x}

\lim_{x \to \infty} \sqrt{4 x} [\sqrt{1 + \sqrt{\frac{1}{4 x} + \sqrt{\frac{1}{{(4 x)}^{3}} + \sqrt{\frac{1}{{(4 x)}^{7}}}}}} - 1] =

letting \sqrt{4 x} = \frac{1}{t} \to {\begin{cases} \frac{1}{4 x} = t^{2} \\ \frac{1}{{(4 x)}^{3}} = t^{6} \\ \frac{1}{{(4 x)}^{7}} = t^{14} \end{cases}

\lim_{t \to 0} \frac{\sqrt{1 + \sqrt{t^{2} + \sqrt{t^{6} + \sqrt{t^{14}}}}} - 1}{t} =

\frac{1}{2} \times \lim_{t \to 0} \frac{\sqrt{t^{2} + \sqrt{t^{6} + t^{7}}}}{t} = \frac{1}{2} \times \lim_{t \to 0} \frac{\sqrt{t^{2} + t^{3} \sqrt{1 + t}}}{t}

= \frac{1}{2} \times \lim_{t \to 0} \frac{t \sqrt{1 + t \sqrt{1 + t}}}{t} = \frac{1}{2}

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