Limx #8594; #8734;x+1 xx+2Rationalising #160;the #160;numerator:limx #8594; #8734;x+2 #160;x+1 x #160;x+1+xx+1+x #8658;limx #8594; #8734;x ...

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Property 4

lim x →∞ x+1-...

Question

$\lim_{x \to \infty} [\{\sqrt{x + 1} - \sqrt{x}\} \sqrt{x + 2}]$

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Solution

$\lim_{x \to \infty} [\{\sqrt{x + 1} - \sqrt{x}\} \sqrt{x + 2}] Rationalising the numerator : \lim_{x \to \infty} [\frac{(\sqrt{x + 2}) \{\sqrt{x + 1} - \sqrt{x}\} \{\sqrt{x + 1} + \sqrt{x}\}}{(\sqrt{x + 1} + \sqrt{x})}] \Rightarrow \lim_{x \to \infty} [\frac{(\sqrt{x + 2}) (x + 1 - x)}{(\sqrt{x + 1} + \sqrt{x})}] Dividing the numerator and the denominator by \sqrt{x} : \lim_{x \to \infty} [\frac{\frac{\sqrt{x + 2}}{\sqrt{x}}}{\frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x}}}] = \lim_{x \to \infty} [\frac{\sqrt{1 + \frac{2}{x}}}{\sqrt{1 + \frac{1}{x}} + 1}] = \frac{1}{\sqrt{1} + 1} = \frac{1}{2}$

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