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:

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

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

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

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

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

   
   

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

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

   
   
   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.

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

   
   
   thus

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