Foundations

What a limit says

A limit is the value a function heads toward as its input closes in on some number. It describes the approach, not the arrival, so the function does not have to be defined at that point for the limit to exist:

\[\lim_{\class{var-gn}{x} \to \class{var-gn}{a}} \class{var-fn}{f}(\class{var-gn}{x}) = \class{var-lim}{L}\]

Reading the notation

\(x \to a\) \(x\) gets arbitrarily close to \(a\) from both sides, never equalling it

\(x \to a^-\) left-hand limit, approaching through values below \(a\)

\(x \to a^+\) right-hand limit, approaching through values above \(a\)

Existence the two-sided limit exists only if both one-sided limits exist and agree both sides must match

\(f(a)\) irrelevant to the limit; it can differ from \(L\) or not exist at all

When a limit fails to exist

Jump the one-sided limits are both finite but different no limit

Infinite the function blows up at a vertical asymptote no limit

Oscillation values swing forever without settling, as \(\sin(1/x)\) does near \(0\) no limit

Removable hole the limit still exists; only the point itself is missing limit exists

A function is continuous at \(a\) when the limit exists, \(f(a)\) exists, and the two are equal. Continuity is what makes the shortcut in the next section legal: for a continuous function, finding the limit is just plugging in.

Limit laws

When two limits both exist, the limit operation passes straight through arithmetic. That lets a complicated expression be broken into pieces and evaluated one at a time:

\[\lim_{x \to a}\big[\class{var-fn}{f}(x) \pm \class{var-gn}{g}(x)\big] = \lim_{x \to a}\class{var-fn}{f}(x) \pm \lim_{x \to a}\class{var-gn}{g}(x)\]

The laws

Constant multiple \(\lim\, c\,f(x) = c \lim f(x)\)

Product \(\lim\, f(x)g(x) = \lim f(x) \cdot \lim g(x)\)

Quotient \(\lim \dfrac{f(x)}{g(x)} = \dfrac{\lim f(x)}{\lim g(x)}\), provided \(\lim g(x) \neq 0\)

Power and root \(\lim [f(x)]^n = [\lim f(x)]^n\), same for \(\sqrt[n]{\ }\)

Composition \(\lim f(g(x)) = f(\lim g(x))\) when \(f\) is continuous there

Direct substitution

Always try it first substitute \(x = a\) and see what comes out step one

Works for polynomials, roots, \(\sin\), \(\cos\), \(e^x\), and \(\ln x\) anywhere on their domain

A number comes out that number is the limit; you are done done

\(\frac{k}{0}\) with \(k \neq 0\) a vertical asymptote; check each side for \(+\infty\) or \(-\infty\) not indeterminate

\(\frac{0}{0}\) indeterminate, so the expression needs rewriting first keep going

Substitution only fails in the cases where the function is not continuous at \(a\), and almost always that failure shows up as \(0/0\). Those are the interesting limits, and every technique below exists to handle them.

Resolving 0/0 by algebra

A \(0/0\) result never means the limit is zero or undefined. It means numerator and denominator share a hidden factor that vanishes at \(a\), and cancelling it exposes the real behavior:

\[\lim_{x \to 4}\frac{x^2 - x - 12}{x - 4} = \lim_{x \to 4}\frac{(x-4)(x+3)}{x-4} = \lim_{x \to 4}(x+3) = 7\]

The four standard moves

Factor and cancel the first thing to try on any rational expression

Multiply by the conjugate use whenever a square root sits inside a difference

Clear inner fractions multiply top and bottom by the common denominator

Divide by the highest power the move for \(x \to \pm\infty\), not for \(x \to a\) end behavior

Limits at infinity of a rational function

Top degree \(<\) bottom limit is \(0\); the denominator wins

Degrees equal limit is the ratio of the leading coefficients

Top degree \(>\) bottom limit is \(\pm\infty\); the numerator wins no finite limit

Growth ranking \(\ln x \ll x^n \ll e^x \ll x!\), which settles most \(\infty/\infty\) races

Cancelling \((x-4)\) is legal because the limit only cares about \(x\) values near \(4\), never \(4\) itself, and there the factor is nonzero. The original function still has a hole at \(x = 4\); the limit fills it in.

Indeterminate forms

An indeterminate form is a result that carries no information. Two competing behaviors are pulling in opposite directions, and which one wins depends entirely on the specific functions, so the expression has to be rewritten before it can be evaluated.

The seven indeterminate forms

\(\dfrac{0}{0}\) and \(\dfrac{\infty}{\infty}\) quotient forms; l'hôpital applies directly ready to use

\(0 \cdot \infty\) a product form; rewrite one factor as a reciprocal

\(\infty - \infty\) combine over a common denominator or rationalize

\(1^\infty\), \(0^0\), \(\infty^0\) exponential forms; take a logarithm first log trick

Forms that only look indeterminate

\(\dfrac{k}{0}\), \(k \neq 0\) grows without bound, so the answer is \(\pm\infty\)

\(\dfrac{0}{k}\) and \(\dfrac{k}{\infty}\) both equal \(0\), no work needed

\(\infty + \infty\) and \(\infty \cdot \infty\) both equal \(\infty\); only the difference is ambiguous

\(0^\infty\) equals \(0\); it is \(0^0\) that is indeterminate easy to confuse

Why it matters l'hôpital on a determinate form gives a wrong answer common error

The clearest example of the ambiguity is \(\infty/\infty\). It can come out as \(0\), as any finite number, or as \(\infty\), depending on how fast each piece grows. That is exactly why the form itself tells you nothing.

L'Hôpital's rule

When a quotient evaluates to \(0/0\) or \(\infty/\infty\), comparing how fast the two parts change resolves the race. Differentiate the numerator and denominator separately and take the limit again:

\[\lim_{x \to a}\frac{\class{var-fn}{f}(x)}{\class{var-gn}{g}(x)} = \lim_{x \to a}\frac{\class{var-fn}{f}'(x)}{\class{var-gn}{g}'(x)}\]

Conditions and cautions

Check the form first the limit must actually be \(0/0\) or \(\infty/\infty\) verify every time

Differentiability \(f\) and \(g\) differentiable near \(a\), with \(g'(x) \neq 0\) there

Not the quotient rule top and bottom are differentiated independently common error

Repeat as needed still indeterminate? apply it again, rechecking the form each pass

If it never resolves the rule is inconclusive; go back to algebra or growth ranking know when to stop

Converting the other forms

\(0 \cdot \infty\) write \(f g\) as \(\dfrac{f}{1/g}\) to make it \(0/0\)

\(\infty - \infty\) combine into a single fraction, which lands on \(0/0\)

\(1^\infty\), \(0^0\), \(\infty^0\) set \(y = f(x)^{g(x)}\), find \(\lim \ln y = \lim g \ln f\)

Finish the log trick the answer is \(e^{\lim \ln y}\), not \(\lim \ln y\) easy to forget

Classic result \(\lim\limits_{x \to \infty}\left(1 + \tfrac{1}{x}\right)^x = e\), a \(1^\infty\) form

The rule works because near \(a\) each function is well approximated by its tangent line, and when both pass through zero the constant terms drop out, leaving only the ratio of slopes. That is why it needs a genuine \(0/0\): the shared zero is what makes the approximation exact in the limit.

Key takeaways

Summary

Limit \(\lim\limits_{x \to a} f(x) = L\), the approach value; exists only if both sides agree

Continuity \(\lim\limits_{x \to a} f(x) = f(a)\), which is what licenses direct substitution

Limit laws limits distribute over sums, products, quotients, and powers when each piece exists

Algebraic toolkit factor, conjugate, clear fractions, or divide by the highest power

Indeterminate forms \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), \(\infty - \infty\), \(1^\infty\), \(0^0\), \(\infty^0\)

L'Hôpital \(\lim \frac{f}{g} = \lim \frac{f'}{g'}\), only after confirming \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)

Indeterminate vs. undefined \(\frac{0}{0}\) needs more work; \(\frac{k}{0}\) is already an infinite answer