>Limits>How to Interpret and Calculate Limits of a Function

How to Interpret and Calculate Limits of a Function

Limits are values that say something about what happens to an expression when a variable approaches a certain value. The value of the limit can be $- \infty$ , $\infty$ , or any number on the number line.

We use a specific notation for limits:

\lim_{x \to a} expression

You read this as “the limit of expression as $x$ approaches $a$ ”, or “the limit of expression as $x$ tends to $a$ ”.

Here, “ $\lim$ ” is an abbreviation for “limit”. Mathematicians try to make things as simple and intuitive as possible.

Rule

Limits of Polynomials

For the polynomial

f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{2} x^{2} + a_{1} x + a_{0}

f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{2} x^{2} + a_{1} x + a_{0}

the limit

\lim_{x \to a} f (x)

is equal to the function value $f (a)$ for all $a \in ℝ$ . Therefore,

\lim_{x \to a} f (x) = f (a)

Example 1

Find the limit of $f (x) = x^{2} + 3 x - 4$ when $x \to 3$

\begin{array}{l} \lim_{x \to 3} f (x) & = \lim_{x \to 3} x^{2} + 3 x - 4 \\ = 3^{2} + 3 \cdot 3 - 4 \\ = 9 + 9 - 4 \\ = 14 \end{array}

Let $f (x)$ be a polynomial of degree $n$ . Thus

f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{2} x^{2} + a_{1} x + a_{0}

f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{2} x^{2} + a_{1} x + a_{0}

x

approaches plus or minus infinity, then

a_{n} x^{n}

eventually becomes the dominant term in the expression, and it will determine whether the function value of

f (x)

is positive or negative.

Consider the limit of $f (x)$ when $x \to \infty$ . The dominant term affects the sign of the function value of $f (x)$ , since $x^{n}$ tends to positive infinity if $n$ is an even number, and it tends to negative infinity if $n$ is an odd number. Below is a summary of the different cases:

Rule

Limits of Polynomials When $x$ Tends to Plus/Minus Infinity

For $a_{n} > 0$ and $n$ an even number, you’ve got

\begin{matrix} \lim_{x \to \infty} f (x) = \infty \\ and \\ \lim_{x \to - \infty} f (x) = \infty \end{matrix}

\lim_{x \to \infty} f (x) = \infty and \lim_{x \to - \infty} f (x) = \infty

For

a_{n} < 0

and

n

an even number, you’ve got

\begin{matrix} \lim_{x \to \infty} f (x) = - \infty \\ and \\ \lim_{x \to - \infty} f (x) = - \infty \end{matrix}

\lim_{x \to \infty} f (x) = - \infty and \lim_{x \to - \infty} f (x) = - \infty

For

a_{n} > 0

and

n

an odd number, you’ve got

\begin{matrix} \lim_{x \to \infty} f (x) = \infty \\ and \\ \lim_{x \to - \infty} f (x) = - \infty \end{matrix}

\lim_{x \to \infty} f (x) = \infty and \lim_{x \to - \infty} f (x) = - \infty

For

a_{n} < 0

and

n

an odd number, you’ve got

\begin{matrix} \lim_{x \to \infty} f (x) = - \infty \\ and \\ \lim_{x \to - \infty} f (x) = \infty \end{matrix}

\lim_{x \to \infty} f (x) = - \infty and \lim_{x \to - \infty} f (x) = \infty

In any case, a polynomial tends to either plus or minus infinity when

x \to \pm \infty

Example 2

Find the limit of $\lim_{x \to - \infty} - 2 x^{3} - 2 x + 5$

Here $f (x)$ is a polynomial of degree 3 and thus of odd degree. In addition, the coefficient of the highest degree term is negative. The limit is therefore

\lim_{x \to - \infty} - 2 x^{3} - 2 x + 5 = \infty

Rule

Trick for Finding Limits of Rational Functions

If both the numerator and the denominator in a fraction tend to zero when $x \to a$ , you can factorize the numerator and denominator separately to find the limit of the fraction $\lim_{x \to a} \frac{f (x)}{g (x)}$ .
If both the numerator and the denominator in a fraction tend to infinity when $x \to a$ , you can divide all the terms in the expression by the highest power of $x$ in the expression.

Note! When you have a simple fraction $\frac{1}{x}$ and $x \to \infty$ , then $\frac{1}{x} = 0$ . All fractions where $x$ is only found in the denominator approach 0 when $x$ approaches infinity. Neat!

Rule

Some Important Limits

When $a \in ℝ ∖ {0}$ , the following are true:

\begin{matrix} \lim_{x \to \infty} \frac{a}{x} = 0 \\ \lim_{x \to - \infty} \frac{a}{x} = 0 \\ \lim_{x \to 0} \frac{a}{x} = \infty \end{matrix}

\lim_{x \to \infty} \frac{a}{x} = 0 \lim_{x \to - \infty} \frac{a}{x} = 0 \lim_{x \to 0} \frac{a}{x} = \infty

Note! $\infty$ is not a number! So no matter how big a number you choose on the number line, $\infty$ is infinitely larger. This means that the ratio between $x$ and $a$ when $x \to \infty$ or $x \to - \infty$ is always infinitely large.

Example 3

Find the limit of $\lim_{x \to 1} \frac{x^{2} - x}{x - 1}$

You then have

\lim_{x \to 1} \frac{x^{2} - x}{x - 1} = \lim_{x \to 1} \frac{x (x-1)}{x-1} = \lim_{x \to 1} x = 1

\lim_{x \to 1} \frac{x^{2} - x}{x - 1} = \lim_{x \to 1} \frac{x (x-1)}{x-1} = \lim_{x \to 1} x = 1

Example 4

Find the limit of $\lim_{x \to \infty} \frac{2 x^{2} - 3 x}{x^{2} + 2}$

You then have

\begin{array}{l} \lim_{x \to \infty} \frac{2 x^{2} - 3 x}{x^{2} + 2} & = \lim_{x \to \infty} \frac{\frac{2 x^{2}}{x^{2}} - \frac{3 x}{x^{2}}}{\frac{x^{2}}{x^{2}} + \frac{2}{x^{2}}} \\ = \lim_{x \to \infty} \frac{2 - \frac{3}{x}}{1 + \frac{2}{x^{2}}} \\ = \frac{2 - 0}{1 + 0} \\ = 2 \end{array}

Example 5

Find the limit of $\lim_{x \to 0} \frac{x^{2} - 2 x}{x}$

You then have

\begin{array}{l} \lim_{x \to 0} \frac{x^{2} - 2 x}{x} & = \lim_{x \to 0} \frac{x (x - 2)}{x} \\ = \lim_{x \to 0} \frac{x (x - 2)}{x} \\ = \lim_{x \to 0} x - 2 \\ = - 2 \end{array}

Rule

Limits of Exponential Functions

For $0 < a < 1$ you’ve got

\lim_{x \to \infty} a^{x} = 0 and \lim_{x \to - \infty} a^{x} = \infty

For $a > 1$ you’ve got

\lim_{x \to \infty} a^{x} = \infty and \lim_{x \to - \infty} a^{x} = 0

For $a = 1$ you’ve got

\lim_{x \to \infty} 1^{x} = \lim_{x \to - \infty} 1^{x} = 1

Example 6

Look at the function $f (x) = e^{0.5 x}$ . Is the function value approaching zero or infinity when $x \to \infty$ ?

Here, you must rewrite the function to find the growth factor, since it’s not obvious whether the growth factor is greater or less than 1:

e^{0.5 x} = {(e^{0.5})}^{x} \approx 1.64 8^{x}

Thus, you see that

\lim_{x \to \infty} f (x) = \infty

Note! You could have seen that $e^{0.5} > 1$ since $e^{x}$ is a strictly increasing function and $e^{0} = 1$ .

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How Do You Determine Continuity of a Function?

Go back

Limits of Polynomials

Limits of Polynomials When x Tends to Plus/Minus Infinity

Trick for Finding Limits of Rational Functions

Some Important Limits

Limits of Exponential Functions

Limits of Polynomials When $x$ Tends to Plus/Minus Infinity