# Limit involving square root

I'd like to know the value of the following limit $$\lim_{t \rightarrow +\infty} \{ (t+1)^{\frac{1}{2}} - (t)^{\frac{1}{2}} \} = ?$$

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Multiplicate.  – Did Feb 12 '12 at 13:39
Try multiplying by $\frac{\sqrt{t + 1} + \sqrt{t}}{\sqrt{t + 1} + \sqrt{t}}$ – Pedro Feb 12 '12 at 14:33

Firstly, this is a typical home work question.

$$\lim_{t \to \infty}{\sqrt{t+1}-\sqrt{t}}=\lim_{t \to \infty}\dfrac{1}{\sqrt{t}(\sqrt{1+\frac{1}{t}}+1)}$$

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thanks................ – user24802 Feb 12 '12 at 14:22

Problems of this type, can usually be handled by multiplying and dividing the given expression by the "conjugate of the square root part" and using the formula $(a-b)(a+b)=a^2-b^2$:

\eqalign{ (t+{\textstyle{1 }})^{1/2} -t^{{1/2} }&= \bigl((t+{\textstyle{1 }})^{1/2} -t^{{1/2}}\bigr)\cdot \underbrace{{(t+{\textstyle{1 }})^{1/2} +t^{{1/2}}\over (t+{\textstyle{1 }})^{1/2} +t^{{1/2}}}}_{=1}\cr &= {\bigl((t+{\textstyle{1 }})^{1/2} -t^{{1/2}}\bigr) \bigl( (t+{\textstyle{1 }})^{1/2} +t^{{1/2}} \bigr) \over(t+{\textstyle{1 }})^{1/2} +t^{{1/2}}}\cr &={ (t+{1 })-t \over(t+{\textstyle{1 }})^{1/2} +t^{{1/2}}}\cr &={ 1 \over (t+{\textstyle{1 }})^{1/2} +t^{{1/2}}}.\cr }

You should be able to take the limit now...

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thanks..................... – user24802 Feb 12 '12 at 14:21