Limit of a Function at a Point

Approaching a Point from Both Sides

Look at $f(x) = \dfrac{x^2-1}{x-1}$ near $x = 1$. The function is not defined at $x = 1$ (the denominator is $0$). But you can get as close to $1$ as you like without touching it, from the left and from the right. For $x \ne 1$ you can write $\dfrac{x^2-1}{x-1} = x + 1$, so the graph is the straight line $y = x + 1$ with a hole at $(1, 2)$.

The key words are "but never equals $a$." The limit only cares about what happens near $a$.

A table of values gives evidence for a limit, not a proof. It works well when the values from the left and the right visibly settle on the same number, as in this table:

One-Sided Limits

Splitting the approach into "from the left" and "from the right" is the key tool for piecewise functions and graphs with jumps.

The reason is simple: the limit asks $f(x)$ to approach one single number from every direction. If the two sides disagree, there is no single number.

Reading limits from a graph

On a graph, follow the curve toward $x = a$ with your finger from the left, and note the height you are heading to. Then do the same from the right. An open circle means the curve gets there but the point is not included. A filled dot is the actual value $f(a)$.

Here $\displaystyle\lim_{x\to2^-}f(x)=3$ and $\displaystyle\lim_{x\to2^+}f(x)=4$. The one-sided limits are different, so $\displaystyle\lim_{x\to2}f(x)$ does not exist. The filled dot $f(2)=2$ plays no role in the limit.

Absolute value and one-sided limits

The absolute value changes its formula at the point where its inside is zero, so you must split into one-sided limits there. Remember $|u| = u$ if $u>0$ and $|u| = -u$ if $u<0$.

The Limit vs. the Value, and Direct Substitution

$\displaystyle\lim_{x\to a} f(x)$ and $f(a)$ are two different questions. The first asks what happens around $a$. The second asks what happens exactly at $a$. They can agree, disagree, or $f(a)$ may not exist.

The curve heads to the open circle from both sides, so that height is the limit. The filled dot is the real value $f(a)$, at a different height. The function "jumps" there, yet the limit is still perfectly well defined.

When a function is nice at $a$ (no division by zero, no negative inside a square root, and so on), the limit and the value agree. That gives the fastest method.

Indeterminate Form 0/0: Factor and Cancel

When substituting gives $\frac{0}{0}$, both the numerator and the denominator contain the factor $(x-a)$. Factor, cancel it, then substitute again.

Why is canceling allowed? A limit never looks at $x = a$ itself. For every $x \ne a$, the fraction $\frac{x-a}{x-a}$ equals $1$, so the simplified function has the same values as the original everywhere near $a$.

For polynomials of degree 3 or more, use synthetic division to find the factor $(x-a)$. Some problems call this Ruffini's rule: it is the same method.

Radicals: Multiply by the Conjugate

When a square root causes the $\frac{0}{0}$, factoring doesn't help. Instead, use the conjugate: the same two terms with the opposite sign in the middle. The conjugate of $\sqrt{x}-3$ is $\sqrt{x}+3$.

Why does it work? Because $(A-B)(A+B) = A^2-B^2$, the square root disappears and a factor $(x-a)$ appears that you can cancel.

When there are roots in both the numerator and the denominator, use the conjugate of each.

More Indeterminate Forms: Combine Fractions and Use Identities

Two more situations show up often.

Difference of fractions ($\infty-\infty$)

Near a point where both fractions blow up, the difference looks like $\infty-\infty$. Combine them over a common denominator, then simplify.

Trigonometric limits: use identities

If substitution gives $\frac{0}{0}$ with $\sin$ and $\cos$, use the Pythagorean identity $\sin^2x+\cos^2x=1$ to build a factor you can cancel. In particular, $1-\cos^2x=\sin^2x$ and $1-\sin^2 x=\cos^2 x$, and each one is a difference of squares.

Infinite Limits and Vertical Asymptotes

Sometimes, as $x$ approaches $a$, the values $f(x)$ don't settle on a number. They grow without bound (or drop without bound). That is also information about the limit.

When substitution gives $\dfrac{k}{0}$ with $k\ne0$, find the sign of the denominator on each side.

If the denominator keeps the same sign on both sides, the infinities agree. For example, $\displaystyle\lim_{x\to-2}\dfrac{3}{(x+2)^2}=+\infty$, because $(x+2)^2$ is a tiny positive number on both sides.

Logarithms

The graph of $\ln u$ falls without bound as $u$ gets close to $0$ from the right: $\ln 0.1\approx-2.3$, $\ln 0.001\approx-6.9$, and so on. So $\displaystyle\lim_{u\to0^+}\ln u=-\infty$. For $\ln(x-3)$ with $x\to3^+$, the inside $x-3$ approaches $0$ through positive values, so the limit is $-\infty$.

Finite, infinite, or does not exist?

When you read a graph, there are three possible outcomes at $x=a$. If the two sides head to the same height, the limit is that finite number. If both sides go up (or both go down) forever, the limit is $+\infty$ (or $-\infty$). If the sides disagree (a jump, or one up and one down), the limit does not exist.

Finding Unknown Parameters

Many problems give a function with an unknown constant, such as $k$, $a$ or $b$, and ask you to choose it so that a limit exists. There are two situations.

Situation 1: a piecewise function. At each point where the formula changes, the left and right limits must be equal. That gives one equation per point.

If there are two change points, you get two equations. Solve the system by substitution or elimination. If the problem also asks the limit to equal $f(a)$, include that value too, as an additional equation.

Situation 2: a $\frac{0}{0}$ form with a parameter. If the denominator approaches $0$ and the limit is to be a finite number, the numerator must also approach $0$. Otherwise you get $\frac{k}{0}$ and the limit is infinite. So set the numerator equal to $0$ at $x=a$.

Limits in Context

Limits answer a practical question: "What value does this quantity approach?" even when the formula can't be evaluated at the exact input.

Always finish with a sentence and units. The tools are the same as before: factor, use a conjugate, or combine fractions.

Secant Slopes and Instantaneous Rates

This is where limits earn their keep. The average rate of change of $s$ over a short interval is $\dfrac{\Delta s}{\Delta t}$, which is also the slope of a secant line. The instantaneous rate is the limit of that quotient as the interval shrinks to zero. The quotient is always $\frac{0}{0}$ at the end, so you need the algebra from the previous sections.

Back to the start: the speedometer reading is the limit of average speeds over shorter and shorter time intervals. At the exact instant the quotient looks like $\frac{0}{0}$, but its limit is a perfectly good number. The same idea gives the slope of the tangent line to a curve: it is the limit of the slopes of secant lines.

Summary