I'd like to know the value of the following limit $ \lim_{t \rightarrow +\infty} \{ (t+1)^{\frac{1}{2}} - (t)^{\frac{1}{2}} \} = ? $
Limit involving square root
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0Try multiplying by $\frac{\sqrt{t + 1} + \sqrt{t}}{\sqrt{t + 1} + \sqrt{t}}$ – 2012-02-12
2 Answers
Firstly, this is a typical home work question.
$\lim_{t \to \infty}{\sqrt{t+1}-\sqrt{t}}=\lim_{t \to \infty}\dfrac{1}{\sqrt{t}(\sqrt{1+\frac{1}{t}}+1)}$
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...