Evaluating Limits

Use direct substitution when continuity at the approach point is known, as explained in Paul's limit calculations. A defined value alone does not justify substitution: a function equal to $0$ away from $a$ and $1$ at $a$ has limit $0$, not $1$. If substitution gives an indeterminate form, transform the expression without changing it on a punctured neighborhood.

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 both the numerator and denominator of 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.

For a squeeze example, $-x^2\le x^2\sin(1/x)\le x^2$ for $x\ne0$, so the limit at $0$ is $0$ despite the oscillating factor. A numerical table may suggest this result but cannot prove it.

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.

L'Hôpital's rule belongs after differentiation rules. It must not be used to prove a basic limit if the derivative formula used in the proof was itself derived from that limit. Nor does failure of the derivative quotient to converge prove that the original limit fails.

3Blue1Brown’s explanation of L’Hôpital’s rule compares the small changes in numerator and denominator near a shared zero. Use the visual argument to understand the derivative ratio, then check the hypotheses above before applying it.

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