Calculating Limits

Direct substitution

The first thing to try is always the simplest: plug in $x = a$. For most of the functions you know, that is all you need.

After you substitute, the result tells you what to do next.

Limits at infinity of polynomials and rational functions

When $x$ is huge (positive or negative), the term with the highest power swamps all the others. That is the whole idea.

For a quotient of polynomials, compare degrees.

The graph flattens out toward the horizontal asymptote $y = \frac{4}{3}$ on both ends.

Exponential and logarithmic limits

Exponentials and logarithms also have clear long-run behavior.

Logarithm properties turn a messy difference into one clean quotient.

One-sided limits and reading graphs

Sometimes the function behaves differently on the two sides of $a$. We write $\lim_{x\to a^-} f(x)$ for the limit as $x$ approaches $a$ from the left (values less than $a$) and $\lim_{x\to a^+} f(x)$ for the limit from the right.

A quotient with a denominator that goes to $0$ (and a numerator that does not) blows up. Check the sign on each side.

Piecewise functions are the other classic case: use the piece that applies on each side.

Reading limits from a graph works the same way: follow the curve toward $x=a$ from each side and note the height it is heading toward. Ignore the actual point at $x=a$: an open circle or a hole does not change a limit. For $x\to+\infty$, look at where the right end of the curve levels off (a horizontal asymptote), or whether it keeps rising or falling.

Indeterminate forms and factoring

Here is the full list of forms that substitution cannot settle:

When a quotient of polynomials gives $\frac{0}{0}$ at $x=a$, both polynomials have a root at $a$. By the Factor Theorem, $(x-a)$ divides both. Factor, cancel, then substitute.

On the graph, the function has a hole at $x=-2$ (the limit exists, but the function is undefined there) and a vertical asymptote at $x=2$.

A related situation: finding a parameter so that a limit exists.

Conjugates with radicals

When the $\frac{0}{0}$ (or $\infty-\infty$) comes with a square root, factoring does not work. Multiply by the conjugate instead. Since $(a-b)(a+b)=a^2-b^2$, the conjugate turns the root into a plain expression.

The same trick works at infinity for $\infty-\infty$ with a root. Add one more move: divide the top and bottom by the highest power of $x$.

Handling $\infty-\infty$

You have a few strategies. Pick the one that matches the expression:

• Two fractions: combine them over a common denominator.

• A root minus something: multiply by the conjugate (the previous section).

• Roots with different indices: split the expression and treat each piece separately.

Two limits can finish a problem that has an unknown parameter. For a finite limit, the leading terms must cancel.

When the roots have different indices, subtract and add $x$ so that each piece is a standard $\infty-\infty$ you can handle on its own. For a cube root use $A-B=\dfrac{A^3-B^3}{A^2+AB+B^2}$.

For an expression like $\sqrt[3]{x^3+6x^2}-\sqrt{x^2+2x}$, write it as $\left(\sqrt[3]{x^3+6x^2}-x\right)-\left(\sqrt{x^2+2x}-x\right)$ and find the limit of each bracket.

Special limits and simplifying first

Some limits involving trig functions cannot be reached by algebra alone. You need one known limit.

The graph has a hole at $x=0$, but the curve clearly heads to $1$ from both sides.

To use it, force the argument of the sine to match the denominator.

The Pythagorean identity gives a second move for expressions with $1-\cos x$: multiply by the conjugate $1+\cos x$, since $(1-\cos x)(1+\cos x)=1-\cos^2x=\sin^2x$.

Another habit worth building: when a quantity is a product, such as an area, multiply it out and simplify before you substitute.

(The height itself is a $\frac{0}{0}$ form. Multiplying by the conjugate shows $h(x)\to\frac{1}{4}$. The rectangle's area goes to $0$ because its width shrinks while the height stays finite.) A right triangle works the same way, with area $\frac{1}{2}\cdot\text{base}\cdot\text{height}$.

The $1^\infty$ form and the number $e$

If the base of a power goes to $1$ and the exponent goes to infinity, the answer is usually a power of $e$. Recall that $e=\lim_{u\to\infty}\left(1+\frac{1}{u}\right)^u\approx 2.718$.

Sometimes $g-1$ needs more algebra, because you must subtract the two polynomials.

If the problem has an unknown parameter and tells you the final answer, set $L$ equal to the exponent of $e$ you are given and solve for the parameter.

Limits in context

In an application, a limit at infinity describes the long run, and a limit at a blow-up describes what happens at a boundary. Always finish by saying what the number means.

Back to the start: the average cost never reaches $\$8$ (the setup fee always adds a little), but it gets as close as you like. That is what the horizontal asymptote $y=8$ says.

Here is how to read a few common results:

• A limit at infinity that is a number $L$ is a steady state or a ceiling (a horizontal asymptote $y=L$).

• A limit of $0$ means the quantity fades away (medicine leaving the body).

• A limit of $+\infty$ means the quantity grows without bound, which is often a warning that the model is impossible or the situation is not sustainable.

• A one-sided limit such as $v\to c^-$ matters when the variable can only approach a boundary from one side.

Summary