Evaluating Limits

Start with direct substitution. If it produces a finite value in the domain, the problem is usually finished. If it produces an indeterminate form, choose a transformation that preserves the function near the point of approach.

Factor and Cancel

For $x\ne3$,

$$\frac{x^2-9}{x-3}=x+3,$$

so

$$\lim_{x\to3}\frac{x^2-9}{x-3}=6.$$

Cancellation is valid here because a limit only requires agreement in a punctured neighborhood; the original expression may remain undefined at $x=3$.

Rationalize a Radical

Multiplying by a conjugate exposes a cancellable factor:

$$\begin{aligned} \lim_{x\to9}\frac{3-\sqrt{x}}{x-9} &=\lim_{x\to9}\frac{(3-\sqrt{x})(3+\sqrt{x})}{(x-9)(3+\sqrt{x})}\\ &=\lim_{x\to9}\frac{-1}{3+\sqrt{x}}\\ &=-\frac16. \end{aligned}$$

Compare Leading Terms at Infinity

Divide a rational function by the largest power in its denominator:

$$\lim_{x\to\infty}\frac{3x^2-4}{5x-2x^2} =\lim_{x\to\infty}\frac{3-4/x^2}{5/x-2} =-\frac32.$$

Check Both Sides of a Boundary

For a piecewise function, compute left- and right-hand limits with the formulas active on their respective sides. The two-sided limit exists only when those values agree.

Combine a Difference Quotient

Assuming $x\ne0$,

$$\begin{aligned} \lim_{h\to0}\frac{1}{h} \left(\frac{1}{x+h}-\frac{1}{x}\right) &=\lim_{h\to0}\frac{-1}{x(x+h)}\\ &=-\frac{1}{x^2}. \end{aligned}$$

The factor $1/h$ is essential; without it, the original difference approaches zero.

Use L'Hôpital's Rule Only After Checking Its Hypotheses

For a quotient with the indeterminate form $0/0$ or $\infty/\infty$, L'Hôpital's rule may replace the quotient with $f'(x)/g'(x)$ when the functions are differentiable on an appropriate punctured interval, $g'(x)\ne0$ there, and the derivative quotient has a limit (finite or infinite). It is not a general license to differentiate numerator and denominator separately.