Use the definition of continuity to prove that the function f defined by f(x) = sqrt(x) is continuous at every nonnegative real number.
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Let $\displaystyle (a,b)$ be an open interval in $\displaystyle D = [0, \infty$. Then
$\displaystyle f^{-1}\left((a,b)\right) = \{x \in D : f(x) \in (a,b)\} = \{x \in D : a < \sqrt x < b\}$
$\displaystyle = \{x \in D : a^2 < x < b^2\} = (a^2,b^2)$.
Since the inverse image of every open interval is an open interval, $\displaystyle f$ is continuous on $\displaystyle D$.