Steady State
MAT161 / Lecture 4
Lecture 4 · Barnett ch. 3 to 4 · about 45 min

Limits

What a limit means before any derivative is in sight: the value a function approaches, one-sided limits, the algebra that removes 0/0, and what happens as output grows without bound.

By the end you can
  • Read one-sided limits from a graph and decide when a two-sided limit exists
  • Apply the limit laws to evaluate limits of sums, products and quotients
  • Factor to resolve a 0/0 indeterminate form at a removable discontinuity
  • Evaluate limits at infinity to find horizontal asymptotes, such as long-run average cost

What a limit means

A limit describes what a function approaches as the input gets arbitrarily close to some value, whether or not the function actually reaches that value there. This is a different question from “what is f(a)f(a)?” — a limit can exist even where f(a)f(a) does not, and f(a)f(a) can exist without the limit existing.

Limit notationFormula sheet →
lim⁡x→af(x)=L\lim_{x \to a} f(x) = L
aa
the input value being approached
LL
the number f(x) approaches
x→ax \to a
x gets arbitrarily close to a, from either side, without ever equaling a

Use it when you need to state or compute what a function approaches near a point, especially one where the function itself may misbehave.

One-sided limits

The left-hand limit looks only at inputs approaching aa from below; the right-hand limit, from above. The two-sided limit exists only when both agree.

Existence of a two-sided limitConditionFormula sheet →
lim⁡x→af(x)=L  ⟺  lim⁡x→a−f(x)=lim⁡x→a+f(x)=L\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L
x→a−x \to a^-
approach from the left (below a)
x→a+x \to a^+
approach from the right (above a)

Use it when a piecewise function or a graph has a possible jump, and you need to check whether the limit exists at the seam.

Check your understanding

A piecewise price function jumps from 4 + 1 = 5 (left-hand limit) to 4 − 1 = 3 (right-hand limit) as x→4x \to 4. Does lim⁡x→4f(x)\lim_{x \to 4} f(x) exist?

Limit laws

For well-behaved (continuous) functions, limits can be computed by direct substitution, and they combine the way you would hope.

Limit lawsRulesFormula sheet →
lim⁡x→a[f(x)±g(x)]=lim⁡x→af(x)±lim⁡x→ag(x)lim⁡x→a[f(x) g(x)]=lim⁡x→af(x)⋅lim⁡x→ag(x)\lim_{x \to a}[f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) \qquad \lim_{x \to a}[f(x)\,g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)
f,gf, g
the two functions being combined, each with an individually existing limit

Use it when a limit is built from sums, differences or products of simpler pieces you already know how to evaluate.

For a quotient, the same rule holds provided the limit of the denominator is not zero: lim⁡fg=lim⁡flim⁡g\lim \frac{f}{g} = \frac{\lim f}{\lim g} when lim⁡g≠0\lim g \neq 0. When the denominator’s limit is zero, direct substitution fails and more work is needed — the next section handles exactly this case.

Your turn

Given lim⁡x→3f(x)=8\lim_{x \to 3} f(x) = 8 and lim⁡x→3g(x)=2\lim_{x \to 3} g(x) = 2, find lim⁡x→3[3f(x)−g(x)2]\lim_{x \to 3} [3f(x) - g(x)^2].

Removable discontinuities: the 0/0 form

If direct substitution gives 00\frac{0}{0}, the function is not necessarily undefined in the limit — it usually means a common factor can be cancelled first.

Worked example · Resolving 0/0 by factoring0/5

Find lim⁡x→2x2−4x−2\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}.

Try it: approach from both sides

Interactive · Approaching a limit
f(x)=x2−4x−2f(x) = \dfrac{x^2 - 4}{x - 2}

Blue point approaches from the left, orange from the right, as h shrinks.

hf(a − h)f(a + h)
13.0005.000
0.53.5004.500
0.13.9004.100
0.013.9904.010
0.0013.9994.001

Both sides settle on the same number: lim⁡x→af(x)=3.999\lim_{x \to a} f(x) = 3.999.

Switch between the three cases. The hole shows a limit that exists even though the function is undefined there. The jump shows a limit that fails to exist because the two sides disagree, even though the function itself is perfectly defined at that point. The asymptote shows a limit that fails to exist because the function grows without bound.

Your turn

Find lim⁡x→5x2−25x−5\lim_{x \to 5} \dfrac{x^2 - 25}{x - 5}.

Limits at infinity

Letting xx grow without bound answers questions like: what happens to average cost as output becomes very large? Divide numerator and denominator by the highest power of xx present.

Limits at infinityFormula sheet →
lim⁡x→∞anxn+…bmxm+…={0n<man/bmn=m±∞n>m\lim_{x \to \infty} \frac{a_n x^n + \dots}{b_m x^m + \dots} = \begin{cases} 0 & n < m \\ a_n/b_m & n = m \\ \pm\infty & n > m \end{cases}
n,mn, m
the degrees of the numerator and denominator
an,bma_n, b_m
their leading coefficients

Use it when a rational function's behaviour as x becomes very large is needed, such as long-run average cost.

Worked example · Long-run average cost0/4

A firm's average cost is Cˉ(x)=5000+20xx\bar C(x) = \dfrac{5000 + 20x}{x}. Find lim⁡x→∞Cˉ(x)\lim_{x \to \infty} \bar C(x) and interpret it.

Check your understanding

For lim⁡x→∞3x2+16x+5\lim_{x \to \infty} \dfrac{3x^2 + 1}{6x + 5}, what is the correct classification and value?

Exam practice

Exam question 1

Find lim⁡x→4(2x2−5x+1)\lim_{x \to 4} (2x^2 - 5x + 1).

Exam question 2

Find lim⁡x→−3x2−9x+3\lim_{x \to -3} \dfrac{x^2 - 9}{x + 3}.

Exam question 3

A firm's average cost is Cˉ(x)=8000+15xx\bar C(x) = \dfrac{8000 + 15x}{x}. Find lim⁡x→∞Cˉ(x)\lim_{x \to \infty} \bar C(x).

Exam question 4

A function satisfies lim⁡x→2−f(x)=7\lim_{x \to 2^-} f(x) = 7 and lim⁡x→2+f(x)=7\lim_{x \to 2^+} f(x) = 7, but f(2)=10f(2) = 10. What is lim⁡x→2f(x)\lim_{x \to 2} f(x)?

Summary and review

Review deck · 10 cards0/10 mastered