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The limit definition fails to produce a finite derivative for $f(x)=\sqrt{x}$ at $x=0$. While the formula yields $\frac{1}{2\sqrt{x}}$ for $x>0$, substituting $x=0$ results in division by zero, indicating that the slope of the tangent line is vertical and the derivative is undefined (infinite) at that point.

Conditions: $f(x)=\sqrt{x}$.; The point of evaluation is $x=0$.; The limit is taken as $h \to 0^+$ (since the domain is $x \ge 0$).

The limit definition fails to produce a finite derivative for $f(x)=\sqrt{x}$ at $x=0$. While the formula yields $\frac{1}{2\sqrt{x}}$ for $x>0$, substituting $x=0$ results in division by zero, indicating that the slope of the tangent line is vertical and the derivative is undefined (infinite) at that point.

Conditions: $f(x)=\sqrt{x}$.; The point of evaluation is $x=0$.; The limit is taken as $h \to 0^+$ (since the domain is $x \ge 0$).