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This question already has an answer here:

Compute the limit $\lim\limits_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$

my attempt:

I tried to multiply top and bottom by the conjugate

$$\begin{align} \lim_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}&=\lim_{x\to+\infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-1}\right)\frac{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\ &=\lim_{x\to+\infty}\frac{\left(\sqrt{x+\sqrt{x}}\right)^2-\left(\sqrt{x-1}\right)^2}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\ &=\lim_{x\to+\infty}\frac{(x+\sqrt{x})-(x-1)}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\ &=\lim_{x\to+\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}} \end{align}$$

But I don't know what I can do after this.

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marked as duplicate by Martin Sleziak, Watson, colormegone, Claude Leibovici calculus Jun 17 '16 at 16:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Let's start from your last line: $$\begin{align} \lim_{x\to+\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}} &= \lim \frac{\sqrt x}{\sqrt x} \frac{\frac{1}{\sqrt x} + 1}{\sqrt{1 + \frac{1}{\sqrt x}} + \sqrt{1 - \frac{1}{ x}}} \\ &= \frac{1}{1 + 1} = \frac{1}{2} \end{align}$$ where we note that everywhere we have $\frac{1}{\sqrt x}$, those terms go to $0$ as $x \to \infty$. The method of factoring out the largest element in the numerator and denominator very often works. $\diamondsuit$

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You may just observe that, as $x\to \infty$: $$ \frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\sim \frac{\sqrt{x}\left(1+\frac1{\sqrt{x}}\right)}{\sqrt{x}\left(\sqrt{1+\frac{1}{\sqrt{x}}}\right)+\sqrt{x}\left(\sqrt{1-\frac{1}{x}}\right)}\sim \frac12. $$

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    $\begingroup$ Your first approximation looks to me like it's exact, not an approximation. $\endgroup$ – raptortech97 Apr 2 '15 at 23:16
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    $\begingroup$ @raptortech97 Thank you for your remark. In fact, it's not necessary to modify it, since it is mathematically correct: if 'equality' is true, then it implies 'approximation', and the latter is sufficient here for the conclusion. $\endgroup$ – Olivier Oloa Apr 3 '15 at 5:20
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$$\begin{align}(x + x^{1/2})^{1/2} &= x^{1/2} + \frac 12 x^{-1/2}x^{1/2} -\frac 1 8 x^{-3/2}x+\cdots\\&=\sqrt x+\frac 12-\frac1{8\sqrt x} +\cdots\\ (x-1)^{1/2} &=\sqrt x -\frac 1{2\sqrt x}+\cdots \end{align}$$

therefore $$(x + x^{1/2})^{1/2} - (x-1)^{1/2}=\frac 1 2 + \frac 3{8\sqrt x} + \cdots \rightarrow \frac 12 \text{ as } x \to \infty.$$

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We have $$\frac{1}{2}=\lim_{x\rightarrow\infty}\frac{\sqrt{x}}{2\sqrt{x}\left(\sqrt{\left(1+\frac{1}{\sqrt{x}}\right)}\right)}\leq\lim_{x\rightarrow\infty}\frac{\sqrt{x}}{\sqrt{x\left(1+\frac{1}{\sqrt{x}}\right)}+\sqrt{x}}\leq\lim_{x\rightarrow\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\leq\lim_{x\rightarrow\infty}\frac{1+\sqrt{x}}{2\sqrt{x}}=\frac{1}{2}$$

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Useful inequality chain

Given $a \ge 0$ and $b > -\frac{1}{2}a$:

  $a+b-2\frac{b^2}{a} \le a+b-\frac{b^2}{a+b} = \frac{a^2+2ab}{a+b} \le \sqrt{a^2+2ab} \le a+b-\frac{1}{2}\frac{b^2}{a+b} \le a+b$

Solution

Given $x \ge 1$:

  $\sqrt{x}+\frac{1}{2}-\frac{1}{4\sqrt{x}+2} \le \sqrt{x+\sqrt{x}} \le \sqrt{x}+\frac{1}{2}$.

  $\sqrt{x}-\frac{1}{2\sqrt{x}}-\frac{1}{4x\sqrt{x}} \le \sqrt{x-1} \le \sqrt{x}-\frac{1}{2\sqrt{x}}$.

Therefore $\sqrt{x+\sqrt{x}}-\sqrt{x-1} \to \frac{1}{2}$ as $x \to \infty$.

Notes

This kind of inequalities are useful when we do not want to use asymptotic expansion but otherwise as shown by abel expansion is the most widely applicable.

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