Limits and Continuity

A progressive guide to evaluating limits, classifying discontinuities, establishing continuity, applying the Intermediate Value Theorem, and using the formal epsilon-delta definition.

What a Measures

A describes nearby behavior. In the notation lim⁡x→af(x)=L\lim_{x\to a}f(x)=L, the input approaches aa while the output approaches LL. The function value at the point is a separate issue: f(a)f(a) may equal LL, differ from LL, or be undefined.

For example, consider

f(x)=x2−1x−1.f(x)=\frac{x^2-1}{x-1}.

For inputs other than 11, factor and simplify:

f(x)=(x−1)(x+1)x−1=x+1.f(x)=\frac{(x-1)(x+1)}{x-1}=x+1.

Therefore,

lim⁡x→1x2−1x−1=2.\lim_{x\to 1}\frac{x^2-1}{x-1}=2.

The original formula is undefined at x=1x=1, but the still exists. This is a .

When estimating from a table, use inputs increasingly close to the target from both sides, such as a−0.1a-0.1, a−0.01a-0.01, a+0.01a+0.01, and a+0.1a+0.1. A graph is interpreted in the same way: trace the output toward the target input from the left and from the right.

Takeaway: A concerns what happens near a point, not necessarily what happens at the point.

One-Sided and Two-Sided Behavior

A two-sided depends on agreement between the two directions of approach. The from the left is written lim⁡x→a−f(x)\lim_{x\to a^-}f(x), and the from the right is written lim⁡x→a+f(x)\lim_{x\to a^+}f(x). The relationship is

lim⁡x→af(x)=L⟺lim⁡x→a−f(x)=L and lim⁡x→a+f(x)=L.\lim_{x\to a}f(x)=L \quad\Longleftrightarrow\quad \lim_{x\to a^-}f(x)=L\text{ and }\lim_{x\to a^+}f(x)=L.

For the piecewise function

f(x)={2x+1,x<3, 7−x,x≥3,f(x)= \begin{cases} 2x+1,&x<3,\ 7-x,&x\ge 3, \end{cases}

we obtain

lim⁡x→3−f(x)=7andlim⁡x→3+f(x)=4.\lim_{x\to 3^-}f(x)=7 \quad\text{and}\quad \lim_{x\to 3^+}f(x)=4.

Since the two values differ, lim⁡x→3f(x)\lim_{x\to 3}f(x) does not exist. This is a .

One-sided reasoning is especially important for piecewise definitions and for rational functions whose denominators approach zero. Always analyze each side before claiming that a two-sided exists.

Takeaway: A two-sided exists only when both one-sided limits agree.

Laws and Algebraic Techniques

When the relevant limits exist, limits follow familiar algebraic laws. If lim⁡x→af(x)=L\lim_{x\to a}f(x)=L and lim⁡x→ag(x)=M\lim_{x\to a}g(x)=M, then

  • lim⁡x→a[f(x)+g(x)]=L+M\lim_{x\to a}[f(x)+g(x)]=L+M;

  • lim⁡x→a[f(x)−g(x)]=L−M\lim_{x\to a}[f(x)-g(x)]=L-M;

  • lim⁡x→a[f(x)g(x)]=LM\lim_{x\to a}[f(x)g(x)]=LM;

  • lim⁡x→af(x)/g(x)=L/M\lim_{x\to a}f(x)/g(x)=L/M when M≠0M\ne0;

  • lim⁡x→a[f(x)]n=Ln\lim_{x\to a}[f(x)]^n=L^n for a positive integer nn.

Direct substitution works for polynomials and for rational functions when the denominator does not approach zero. For example,

lim⁡x→2(3x2−5x+1)=3(2)2−5(2)+1=3.\lim_{x\to 2}(3x^2-5x+1)=3(2)^2-5(2)+1=3.

If direct substitution produces an such as 0/00/0, continue with algebra. For

lim⁡x→3x2−9x−3,\lim_{x\to 3}\frac{x^2-9}{x-3},

factor the numerator:

x2−9x−3=(x−3)(x+3)x−3=x+3,x≠3.\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3, \qquad x\ne3.

Thus,

lim⁡x→3x2−9x−3=6.\lim_{x\to 3}\frac{x^2-9}{x-3}=6.

Other methods include rationalizing with a conjugate, combining fractions, using known trigonometric limits, and applying the squeeze theorem.

Takeaway: Direct substitution is a first test, not a complete strategy when it produces an .

Infinite Limits and Vertical Asymptotes

An infinite describes unbounded function values near a finite input. For example,

lim⁡x→2+1x−2=+∞andlim⁡x→2−1x−2=−∞.\lim_{x\to 2^+}\frac{1}{x-2}=+\infty \quad\text{and}\quad \lim_{x\to 2^-}\frac{1}{x-2}=-\infty.

The line x=2x=2 is therefore a . Infinity here describes unbounded growth; it is not an ordinary real number that the function reaches.

To determine signs near a denominator zero, inspect the factors. For

f(x)=x+1(x−2)(x+3),f(x)=\frac{x+1}{(x-2)(x+3)},

near x=2x=2, the factors x+1x+1 and x+3x+3 are positive. The sign is controlled by x−2x-2: it is negative from the left and positive from the right. Hence,

lim⁡x→2−f(x)=−∞andlim⁡x→2+f(x)=+∞.\lim_{x\to2^-}f(x)=-\infty \quad\text{and}\quad \lim_{x\to2^+}f(x)=+\infty.

Do not confuse this situation with a at infinity. The expressions lim⁡x→af(x)=±∞\lim_{x\to a}f(x)=\pm\infty and lim⁡x→±∞f(x)=L\lim_{x\to\pm\infty}f(x)=L describe different kinds of behavior.

Takeaway: Analyze factor signs on each side of a finite denominator zero to determine infinite behavior.

End Behavior and Horizontal Asymptotes

A at infinity describes end behavior as the input becomes arbitrarily large or arbitrarily negative. For a rational function

f(x)=anxn+⋯bmxm+⋯,f(x)=\frac{a_nx^n+\cdots}{b_mx^m+\cdots},

compare the degrees of the numerator and denominator:

  • If the numerator degree is less than the denominator degree, the is 00.

  • If the degrees are equal, the is the ratio of leading coefficients, an/bma_n/b_m.

  • If the numerator degree is greater, there is no finite ; polynomial division may reveal another asymptotic form.

For example,

lim⁡x→∞5x3−2x+12x3+7x2−4=52.\lim_{x\to\infty}\frac{5x^3-2x+1}{2x^3+7x^2-4}=\frac{5}{2}.

Dividing by x3x^3 gives

5−2/x2+1/x32+7/x−4/x3.\frac{5-2/x^2+1/x^3}{2+7/x-4/x^3}.

As xx approaches infinity, the negative powers of xx approach zero, leaving 5/25/2. Thus, y=5/2y=5/2 is the corresponding .

Takeaway: Limits near a finite input describe local behavior; limits at infinity describe the ends of the graph.

at Points and on Intervals

To establish at x=ax=a, verify all three conditions:

  1. f(a)f(a) is defined.

  2. lim⁡x→af(x)\lim_{x\to a}f(x) exists.

  3. lim⁡x→af(x)=f(a)\lim_{x\to a}f(x)=f(a).

Polynomials are continuous at every real input. Rational functions are continuous wherever their denominators are nonzero, and sums, products, quotients, and compositions preserve wherever they are defined.

For a piecewise function, check the boundary separately. Let

f(x)={x2+1,x<2, ax+3,x≥2.f(x)= \begin{cases} x^2+1,&x<2,\ ax+3,&x\ge2. \end{cases}

The left-hand is

lim⁡x→2−f(x)=22+1=5.\lim_{x\to2^-}f(x)=2^2+1=5.

The right-hand and function value are both 2a+32a+3. requires

5=2a+3,5=2a+3,

so a=1a=1. With this value, both one-sided limits and the function value equal 55.

On a closed interval [a,b][a,b], in the interior is combined with right-hand at aa and left-hand at bb:

lim⁡x→a+f(x)=f(a),lim⁡x→b−f(x)=f(b).\lim_{x\to a^+}f(x)=f(a), \qquad \lim_{x\to b^-}f(x)=f(b).

Takeaway: A convincing justification checks definitions and boundary behavior rather than relying only on the appearance of a formula.

The

The applies when a function is continuous on an entire closed interval. If ff is continuous on [a,b][a,b], then every number between f(a)f(a) and f(b)f(b) occurs at least once for some c∈[a,b]c\in[a,b].

For root existence, it is enough to find opposite endpoint signs. Let

f(x)=x3+x−1.f(x)=x^3+x-1.

This polynomial is continuous on [0,1][0,1], and

f(0)=−1,f(1)=1.f(0)=-1, \qquad f(1)=1.

Because zero lies between the endpoint values, the theorem guarantees a number c∈(0,1)c\in(0,1) such that

f(c)=0.f(c)=0.

This proves that the equation x3+x−1=0x^3+x-1=0 has at least one solution in the interval. It does not prove that the solution is unique or provide its exact value.

The hypothesis is essential. Also, endpoint values alone do not rule out an interior root: for f(x)=(x−1)2f(x)=(x-1)^2 on [0,2][0,2], both endpoint values are 11, but f(1)=0f(1)=0.

A strong IVT argument should explicitly state the interval, verify , calculate endpoint values, identify the intermediate value, and conclude only what the theorem guarantees.

Takeaway: The theorem guarantees existence under , not uniqueness or an exact solution.

The Formal

The formal makes the idea of approaching precise. The statement

lim⁡x→af(x)=L\lim_{x\to a}f(x)=L

means that for every ε>0\varepsilon>0, there exists a δ>0\delta>0 such that

0<∣x−a∣<δ⟹∣f(x)−L∣<ε.0<|x-a|<\delta \quad\Longrightarrow\quad |f(x)-L|<\varepsilon.

The quantity ε\varepsilon specifies the allowed output error, while δ\delta specifies how close the input must be to aa. The condition 0<∣x−a∣0<|x-a| excludes the point itself because a concerns nearby inputs.

To prove

lim⁡x→3(2x+1)=7,\lim_{x\to3}(2x+1)=7,

start with the output distance:

∣(2x+1)−7∣=2∣x−3∣.|(2x+1)-7|=2|x-3|.

Choose

δ=ε2.\delta=\frac{\varepsilon}{2}.

Then 0<∣x−3∣<δ0<|x-3|<\delta implies

∣(2x+1)−7∣=2∣x−3∣<2δ=ε.|(2x+1)-7|=2|x-3|<2\delta=\varepsilon.

Therefore the follows from the definition.

For a linear function f(x)=mx+bf(x)=mx+b,

∣f(x)−f(a)∣=∣m∣∣x−a∣.|f(x)-f(a)|=|m||x-a|.

If m≠0m\ne0, choose δ=ε/∣m∣\delta=\varepsilon/|m|; if m=0m=0, any positive δ\delta works. This confirms that every linear function is continuous everywhere.

Takeaway: In a formal proof, choose an input tolerance that forces the desired output tolerance.

A Reliable Problem-Solving Checklist

Use the following sequence to organize most and problems:

  1. Identify the approach: a finite input, a one-sided approach, or an infinite input.

  2. Try direct substitution.

  3. If the result is an such as 0/00/0, simplify algebraically or use another appropriate method.

  4. For one-sided behavior, analyze the two sides separately.

  5. For rational functions at infinity, compare the numerator and denominator degrees or divide by the highest power of xx.

  6. For , verify that the function is defined, that the exists, and that the equals the function value.

  7. For the , state on the entire closed interval and evaluate the endpoints.

  8. State precisely what has been proved, using “at least one” when uniqueness has not been established.

Common errors include treating 0/00/0 as a value, assuming f(a)f(a) determines the , checking only one side of a two-sided , confusing infinite limits with limits at infinity, and applying the without verifying .

Final checklist: Identify the type of question, select the relevant definition or theorem, verify its hypotheses, show the essential algebra, and phrase the conclusion with the correct level of certainty.