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$$\lim_{x \to \infty}\frac{ \sqrt {x+1}- \sqrt {x}}{\sqrt {x}- \sqrt {x-1}}=?$$

I've tried to rationalize the denominator and during another attempt the numerator, but nothing seems to work. Squeeze theorem got me even worse places. Help me please!

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  • $\begingroup$ note that for $x \rightarrow \infty$, $\sqrt{x}\sim\sqrt{x\pm1}$ $\endgroup$ – tired Oct 29 '16 at 10:51
  • $\begingroup$ Did you try multiplying numerator and denominators by the conjugated binomials to get rid of all square roots? $\endgroup$ – Arnaud D. Oct 29 '16 at 10:51
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    $\begingroup$ $$\frac{ \sqrt {x+1}- \sqrt {x}}{\sqrt {x}- \sqrt {x-1}}=\frac{ \sqrt {x+1}- \sqrt {x}}{\sqrt {x}- \sqrt {x-1}}\frac{ \sqrt {x+1}+ \sqrt {x}}{\sqrt {x}+ \sqrt {x-1}}\frac{ \sqrt {x}+ \sqrt {x-1}}{\sqrt {x+1}+ \sqrt {x}}=\ldots$$ $\endgroup$ – Did Oct 29 '16 at 10:51
  • $\begingroup$ Or, using $u(x)=\sqrt{x}$ and $u'(x)=\frac12\frac1{\sqrt{x}}$, note that there exists $\xi(x)$ in $(x,x+1)$ and $\eta(x)$ in $(x-1,x)$ such that $$\frac{\sqrt{x+1}-\sqrt{x}}{ \sqrt {x}-\sqrt {x-1}}=\frac{\frac1{\sqrt{2\xi(x)}}}{\frac1{2\sqrt{\eta(x)}}}=\ldots$$ $\endgroup$ – Did Oct 29 '16 at 10:54
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You can get rid of the differences of square roots by a simple trick : \begin{align}\frac{ \sqrt {x+1}- \sqrt {x}}{\sqrt {x}- \sqrt {x-1}} & =\frac{ \sqrt {x+1}- \sqrt {x}}{\sqrt {x}- \sqrt {x-1}}\cdot\frac{ \sqrt {x+1}+ \sqrt {x}}{\sqrt {x+1}+\sqrt {x}}\cdot\frac{ \sqrt {x}+ \sqrt {x-1}}{\sqrt {x}+ \sqrt {x-1}}\\ &= \frac{x+1-x}{x-(x-1)}\cdot \frac{ \sqrt {x}+ \sqrt {x-1}}{\sqrt {x+1}+ \sqrt {x}}\\ & =\frac{ \sqrt {x}+ \sqrt {x-1}}{\sqrt {x+1}+ \sqrt {x}} \\ & = \dfrac{1+\sqrt{\dfrac{x-1}{x}}}{\sqrt{\dfrac{x+1}{x}}+1},\end{align} and from there the limit should be easier.

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First: multiplication (numerator - denominator) by denominator's conjugate:

$$\frac{\sqrt{x+1}-\sqrt x}{\sqrt x-\sqrt{x-1}}=\frac{\sqrt{x+1}-\sqrt x}{\sqrt x-\sqrt{x-1}}\cdot\frac{\sqrt x+\sqrt{x-1}}{\sqrt x+\sqrt{x-1}}=$$

$$=\frac{\left(\sqrt{x+1}-\sqrt x\right)\left(\sqrt x+\sqrt{x-1}\right)}1$$

Next: multiplication (numerator - denominator) by conjugate of the original numerator:

$$\frac{\left(\sqrt{x+1}-\sqrt x\right)\left(\sqrt x+\sqrt{x-1}\right)\left(\sqrt{x+1}+\sqrt x\right)}{\sqrt{x+1}+\sqrt x}=$$

$$=\frac{\sqrt x+\sqrt{x-1}}{\sqrt{x+1}+\sqrt x}=\frac{1+\sqrt{1-\frac1x}}{\sqrt{1+\frac1x}+1}\xrightarrow[x\to\infty]{}\frac{1+1}{1+1}=1$$

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Just another approach.

Factor $\sqrt x$ to make $$A=\frac{ \sqrt {x+1}- \sqrt {x}}{\sqrt {x}- \sqrt {x-1}}=\frac{\sqrt {x}\left(\sqrt{1+\frac1x}-1\right)}{ \sqrt {x}\left(1-\sqrt{1-\frac1x}\right)}=\frac{\sqrt{1+\frac1x}-1}{1-\sqrt{1-\frac1x}}$$ Now, make $x=\frac 1y$ to make $$A=\frac{\sqrt{1+y}-1}{1-\sqrt{1- y}}$$ and use either Taylor series around $y=0$ or the generalized binomial theorem $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ $$\sqrt{1-y}=1-\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ This makes $$A=\frac{\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right) }{ \frac{y}{2}+\frac{y^2}{8}+O\left(y^3\right)}$$ Simplify and perform the long division to get $$A=1-\frac{y}{2}+O\left(y^2\right)$$ which shows the limit and how it is approached.

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