Definition of the Derivative

You must cancel $h$ because direct substitution of $h=0$ into the difference quotient $\frac{f(x+h)-f(x)}{h}$ results in the indeterminate form $0/0$. Algebraic simplification, such as expansion and cancellation for polynomials or rationalization for radicals, removes the singularity at $h=0$, allowing the limit to be evaluated by direct substitution of the simplified expression.

Conditions: The function $f$ is differentiable at $x$.; The difference quotient initially yields a $0/0$ form when $h=0$.; Algebraic manipulation is valid for nonzero $h$.

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 $x \ge 0$).

To find the derivative of $f(x)=x^2$, substitute into the difference quotient to get $\frac{(x+h)^2-x^2}{h}$. Expanding the numerator yields $2xh+h^2$.

Conditions: $f(x)=x^2$.; $h$ is nonzero during the cancellation step.; The limit is taken as $h \to 0$ after simplification.

To find the derivative of $f(x)=\sqrt{x}$, substitute into the difference quotient $\frac{\sqrt{x+h}-\sqrt{x}}{h}$. Multiply the numerator and denominator by the conjugate $\sqrt{x+h}+\sqrt{x}$ to rationalize the numerator.

Conditions: $f(x)=\sqrt{x}$.; $x > 0$ for the derivative to be finite.; $h$ is nonzero during the algebraic manipulation.

When the finite limit exists, $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. We do not substitute $h=0$ directly into the quotient; we evaluate its limit.

Conditions: The finite limit must exist.; $h$ approaches zero.