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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 $h$.

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 $h$.

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 $f(x)=x^2$, 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.