Difference two square roots.

Possible method: Use the ``conjugate expression'', i.e. the expression with the middle sign reversed.

  1. Take $ f(x)=\sqrt{x+1}-\sqrt{x}$ , and compute its limit at $ +\infty$ . We have:

    $\displaystyle \underset{x \rightarrow +\infty}{\lim} \sqrt{x+1} =\underset{x \rightarrow +\infty}{\lim} \sqrt{x}= +\infty,$    

    thus this is an undeterminate case. We multiply and divide bythe conjugate expression:

    $\displaystyle f(x)=\frac{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})}{\sqrt{x+1}+\sqrt{x}} =\frac {1}{\sqrt{x+1}+\sqrt{x}}.$    

    The numerator is constant and the denomainator has an infinite limit at $ +\infty$ , thus

    $\displaystyle \underset{x \rightarrow +\infty}{\lim} f(x)=0.$    

  2. Take $ f(x)=x-\sqrt(x)$ . Here too we have an undeterminate case; we use the same method as above.

    $\displaystyle f(x)=\frac{(x-\sqrt{x})(x+\sqrt{x})}{x+\sqrt{x}}=\frac {x^2-x}{x+\sqrt{x}}.$    

    We transformed an undeterminate case into another one: both numerator and denominator have an infinite limit at $ +\infty$ . We use now a variant of te method exposed in subsection subsection limit polynomial and rational.

    $\displaystyle f(x)=\frac{x^2 \; \left( 1 -\frac 1x \right) }{x \; \left( 1 + \f...
...ght)}^{\rightarrow 1} }{\underbrace{ 1 + \frac{1}{\sqrt{x}}}_{\rightarrow 1} },$    

    thus

    $\displaystyle \underset{x \rightarrow +\infty}{\lim} f(x)=+\infty.$    

Noah Dana-Picard 2007-12-28