3 WAYS TO SOLVE LIMITS

Factoring and cancelling resolves the limit by simplifying the expression to a form that is defined near the target point. For $x \ne 2$ near 2, the expression $\frac{x^2+3x-10}{x^2-2x}$ factors into $\frac{(x-2)(x+5)}{x(x-2)}$.

Conditions: The limit is as $x \to 2$.; The original expression is undefined at $x=2$ (indeterminate form $0/0$).; Cancellation is valid only for $x \ne 2$.

Clearing the inner fraction resolves the limit by simplifying the complex rational expression into a polynomial form. Near $x=-1$, with $x\ne-1$, multiplying the numerator and denominator of $\frac{3+3x}{1/x+1}$ by $x$ gives $\frac{3x+3x^2}{1+x}$.

Conditions: The limit is as $x \to -1$.; The expression contains an inner fraction $1/x$ in the denominator.; Multiplication by $x$ is valid for $x \ne 0$, and specifically near $x=-1$ where $x \ne 0$.; The original expression is undefined at $x=-1$ (indeterminate form $0/0$).

Multiplying by the conjugate resolves the limit by eliminating the radical in the denominator, which causes the $0/0$ indeterminacy. For $\lim_{x\to5}\frac{x-5}{\sqrt{x+4}-3}$, multiply the numerator and denominator by $\frac{\sqrt{x+4}+3}{\sqrt{x+4}+3}$.

Conditions: The limit is as $x \to 5$.; The expression involves a radical in the denominator causing a $0/0$ form.; The conjugate is $\sqrt{x+4}+3$.

The simplified functions agree with the original expressions near the target point because the algebraic manipulations (factoring, conjugate multiplication, clearing fractions) are equivalences valid for all $x$ in the domain except possibly the target point itself. Limits describe the behavior of a function as the input approaches a value, not the value of the function at that point.

Conditions: The target point is excluded from the domain of the original expression.; The algebraic simplification is valid for all $x$ near the target point ($x \ne$ target).; The limit exists for the simplified function.