Limits and Computational Approach

In Limits (Intuitive Introduction) we guessed limits from tables and graphs. Here we compute them exactly. This lesson covers:

• the limits of two very simple functions, the constant function and ${y=x};$
• the limit laws, which give the limits of more complicated functions from these two;
• algebraic techniques for the cases where the laws cannot be used directly: factoring, conjugates, signs, leading terms.

On this page: Two basic limits · Limit laws · Polynomials · Rational functions · The form 0/0 · Number over 0 · Limits at infinity · Strategy · Common mistakes · Practice

Two Basic Limits

We start with two functions:

• ${f(x)=k},$ the constant function ($k$ is a fixed number); its graph is a horizontal line;

• ${g(x)=x};$ its graph is the straight line through the origin with slope $1.$

The constant function ${f(x)=k}$

Whatever $x$ is, the value of the constant function is $k.$ So, as $x$ approaches any number $a,$ or grows or decreases without bound, the value stays $k\text{:}$

Limit	Examples
${\lim_{x\to a}k=k}$	${\lim_{x\to2}3=3},$   ${\lim_{x\to-2}3=3}$
${\lim_{x\to+\infty}k=k}$	${\lim_{x\to+\infty}3=3},$   ${\lim_{x\to+\infty}0=0}$
${\lim_{x\to-\infty}k=k}$	${\lim_{x\to-\infty}3=3},$   ${\lim_{x\to-\infty}0=0}$

The function ${g(x)=x}$

The value of this function is $x$ itself. If $x$ approaches $a,$ the value approaches $a.$ If $x$ grows without bound, so does the value, and if $x$ decreases without bound, so does the value:

Limit
${\lim_{x\to a}x=a}$
${\lim_{x\to+\infty}x=+\infty}$
${\lim_{x\to-\infty}x=-\infty}$

The Limit Laws

In words: the limit of a sum is the sum of the limits, the limit of a product is the product of the limits, and so on, provided that the separate limits exist (and, for a quotient, that the limit of the denominator is not $0$).

Remark. Laws (a) and (c) are stated for two functions, but they hold for any finite number of functions. If the limits ${\lim f_1(x)},$ ${\lim f_2(x)},$ …, ${\lim f_n(x)}$ all exist, then

${\lim\,[f_1(x)+\ldots+f_n(x)]}={{\lim f_1(x)+\ldots+\lim f_n(x)},}$

${\lim\,[f_1(x)\cdot\ldots\cdot f_n(x)]}={{\lim f_1(x)\cdot\ldots\cdot\lim f_n(x)}.}$

If $f_1,$ $f_2,$ …, $f_n$ are all the same function $f,$ the second formula gives the power law (for ${n=1},$ $2,$ $3,$ …, if ${\lim f(x)}$ exists):

$$\lim\,[f(x)]^n=\left[\lim f(x)\right]^n.$$

Two useful consequences:

A) ${\lim_{x\to a}x^n}={\left[\lim_{x\to a}x\right]^n}={a^n},$   for ${n=1},$ $2,$ $3,$ …;

B) ${\lim\,[k\cdot g(x)]}={\lim k\cdot\lim g(x)}={{k\cdot\lim g(x)},}$ where $k$ is a constant: a constant factor can be taken out of the limit.

Limits of Polynomials

A polynomial is a function of the form

$$p(x)=c_0+c_1x+c_2x^2+\ldots+c_nx^n,$$

where $c_0,$ $c_1,$ …, $c_n$ are constants (numbers).

Example 1. Find ${\lim_{x\to5}(x^2-4x+3)}.$

${\lim_{x\to5}(x^2-4x+3)}={\lim_{x\to5}x^2-4\lim_{x\to5}x+\lim_{x\to5}3}={5^2-4\cdot5+3}={{8}.}$

We used the laws (a), (b), the consequences A and B, and the limit of a constant. The result is the value of the polynomial at $5.$ This is always so:

Theorem. For any polynomial ${p(x)}={c_0+c_1x+\ldots+c_nx^n}$ and any real number $a,$ ${\lim_{x\to a}p(x)}={c_0+c_1a+\ldots+c_na^n}={{p(a)}.}$

Proof. By the sum law, then the constant-factor law B, then A:

${\lim_{x\to a}p(x)}={\lim_{x\to a}c_0+\lim_{x\to a}c_1x}+{\ldots+\lim_{x\to a}c_nx^n}={c_0+c_1\lim_{x\to a}x}+{\ldots+c_n\lim_{x\to a}x^n}={c_0+c_1a+\ldots+c_na^n}={{p(a)}.}$

So the limit of a polynomial at a point is found by substitution.

Limits of Rational Functions

A rational function is a quotient of two polynomials, ${r(x)=\frac{p(x)}{q(x)}}.$ If ${q(a)\ne0},$ the quotient law and the theorem above give

$$\lim_{x\to a}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)}.$$

Example 2.

${\lim_{x\to2}\frac{5x^3+4}{x-3}}={\frac{\lim_{x\to2}(5x^3+4)}{\lim_{x\to2}(x-3)}}={\frac{5\cdot2^3+4}{2-3}}={\frac{44}{-1}}={{-44}.}$

If ${q(a)=0},$ the quotient law cannot be used, and there are two cases:

• ${p(a)=0}$ too: the "form ${\frac{0}{0}}$". The limit can be any number, $+\infty$ or $-\infty,$ or it may not exist. We have to simplify first.

• ${p(a)\ne0}\text{:}$ the "form ${\frac{\text{number}}{0}}$". The values of the fraction grow without bound in absolute value, and the one-sided limits are $+\infty$ or $-\infty.$

The form $0/0\text{:}$ simplify first

Example 3. Find ${\lim_{x\to2}\frac{x^2-4}{x-2}}.$

At ${x=2}$ both the numerator and the denominator are $0.$ But the limit uses only the values ${x\ne2},$ so we may cancel the common factor ${x-2}\text{:}$

${\lim_{x\to2}\frac{x^2-4}{x-2}}={\lim_{x\to2}\frac{(x-2)(x+2)}{x-2}}={\lim_{x\to2}(x+2)}={{4}.}$

The function is not defined at ${x=2}\text{:}$ its graph is the line ${y=x+2}$ with a hole at $(2, 4).$ Still, the limit is $4.$

Example 4.

${\lim_{x\to3}\frac{x^2-5x+6}{x^2-9}}={\lim_{x\to3}\frac{(x-2)(x-3)}{(x-3)(x+3)}}={\lim_{x\to3}\frac{x-2}{x+3}}={{\frac{1}{6}}.}$

If $a$ is a root of both polynomials, both of them are divisible by ${x-a}$ (see Horner's rule), so the factor ${x-a}$ can always be cancelled. If you still get ${\frac{0}{0}},$ cancel ${x-a}$ again.

Example 5 (a square root: multiply by the conjugate). Find ${\lim_{x\to4}\frac{\sqrt{x}-2}{x-4}}.$

Again ${\frac{0}{0}}.$ We multiply the numerator and the denominator by the conjugate ${\sqrt{x}+2}$ and use ${{(\sqrt{x}-2)(\sqrt{x}+2)=x-4}\text{:}}$

${\lim_{x\to4}\frac{\sqrt{x}-2}{x-4}}={\lim_{x\to4}\frac{x-4}{(x-4)(\sqrt{x}+2)}}={\lim_{x\to4}\frac{1}{\sqrt{x}+2}}={{\frac{1}{4}}.}$

A nonzero number over $0\text{:}$ infinite limits

The simplest example is ${\frac{1}{x}}$ at ${x=0}.$ The limits below are suggested by its graph and by the table.

$x$	${\frac{1}{x}}$	Conclusion
1, 0.1, 0.01, 0.001, …	1, 10, 100, 1000, …	${\lim_{x\to0^+}\frac{1}{x}=+\infty}$
−1, −0.1, −0.01, −0.001, …	−1, −10, −100, −1000, …	${\lim_{x\to0^-}\frac{1}{x}=-\infty}$
1, 10, 100, 1000, …	1, 0.1, 0.01, 0.001, …	${\lim_{x\to+\infty}\frac{1}{x}=0}$
−1, −10, −100, −1000, …	−1, −0.1, −0.01, −0.001, …	${\lim_{x\to-\infty}\frac{1}{x}=0}$

For every real number $a,$ the graph of ${g(x)=\frac{1}{x-a}}$ is the graph of ${f(x)=\frac{1}{x}}$ moved $a$ units to the right if ${a\gt0},$ and $|a|$ units to the left if ${a\lt0}.$ Therefore

${\lim_{x\to a^-}\frac{1}{x-a}=-\infty},$   ${\lim_{x\to a^+}\frac{1}{x-a}=+\infty},$   ${\lim_{x\to-\infty}\frac{1}{x-a}=0},$   ${\lim_{x\to+\infty}\frac{1}{x-a}=0}.$

In general, if the numerator approaches a number ${L\ne0}$ and the denominator approaches $0,$ the absolute value of the fraction grows without bound. If the fraction keeps one sign on one side of $a$ (a rational function always does), that sign decides between $+\infty$ and $-\infty$ on that side.

Example 6. Find ${\lim_{x\to4^+}\frac{2-x}{(x-4)(x+2)}}.$

At ${x=4}$ the numerator is ${2-4=-2}$ and the denominator is $0.$ Mark the zero of the numerator ($2$) and the zeros of the denominator ($-2$ and $4$) on the number line. For large $x$ the fraction is negative (its leading terms give ${\frac{-x}{x^2}}={-\frac{1}{x}}$), and the sign changes at each marked point, because each of them comes from a factor to the first power (a factor such as ${(x-2)^2}$ would not change the sign):

For $x$ a little larger than $4,$ the numerator is close to $-2$ (negative), ${x-4}$ is a small positive number and ${x+2}$ is close to $6$ (positive). So the denominator is a small positive number and the fraction is a large negative number:

$$\lim_{x\to4^+}\frac{2-x}{(x-4)(x+2)}=-\infty.$$

For $x$ a little smaller than $4,$ ${x-4}$ is negative, the fraction is positive and ${\lim_{x\to4^-}\frac{2-x}{(x-4)(x+2)}}={+\infty}.$ The one-sided limits are different, so the two-sided limit at $4$ does not exist. The graph has the vertical asymptotes ${x=-2}$ and ${x=4}\text{:}$

Limits at Infinity

Powers of $x$

For ${n=1},$ $2,$ $3,$ …, the power $x^n$ grows without bound as ${x\to+\infty}.$ As ${x\to-\infty},$ an even power is positive and grows without bound, and an odd power is negative and decreases without bound:

${\lim_{x\to+\infty}x^n=+\infty},$   ${n=1},$ $2,$ $3,$ …

$$\lim_{x\to-\infty}x^n=\begin{cases}+\infty, & n=2,\ 4,\ 6,\ \ldots\\ -\infty, & n=1,\ 3,\ 5,\ \ldots\end{cases}$$

Multiplying by a positive constant does not change an infinite limit, and multiplying by a negative constant changes its sign:

${\lim_{x\to+\infty}2x^5=+\infty},$   ${\lim_{x\to+\infty}(-7x^6)=-\infty},$   ${\lim_{x\to-\infty}2x^5=-\infty},$   ${\lim_{x\to-\infty}(-7x^6)=-\infty}.$

For ${n=1},$ $2,$ $3,$ … the power law and the table above give

${\lim_{x\to+\infty}\frac{1}{x^n}}={\left(\lim_{x\to+\infty}\frac{1}{x}\right)^n}={{0},}$   ${\lim_{x\to-\infty}\frac{1}{x^n}}={\left(\lim_{x\to-\infty}\frac{1}{x}\right)^n}={{0}.}$

Near $0$ the graphs of ${y=\frac{1}{x^n}}$ look like this: for even $n$ both branches go up (${\lim_{x\to0}\frac{1}{x^n}=+\infty}$); for odd $n$ the left branch goes down and the right branch goes up, like ${y=\frac{1}{x}}$ (${\lim_{x\to0^-}\frac{1}{x^n}=-\infty},$ ${\lim_{x\to0^+}\frac{1}{x^n}=+\infty}$).

Polynomials at infinity

A polynomial behaves like its term of highest degree as ${x\to+\infty}$ or ${x\to-\infty}.$ More precisely, if ${c_n\ne0},$ then

${\lim_{x\to+\infty}(c_0+c_1x+\ldots+c_nx^n)}={{\lim_{x\to+\infty}c_nx^n},}$

${\lim_{x\to-\infty}(c_0+c_1x+\ldots+c_nx^n)}={{\lim_{x\to-\infty}c_nx^n}.}$

Why? Factor out $x^n\text{:}$

${c_0+c_1x+\ldots+c_nx^n}={{x^n\left(\frac{c_0}{x^n}+\ldots+\frac{c_{n-1}}{x}+c_n\right)}.}$

As ${x\to\pm\infty},$ every fraction in the brackets approaches $0,$ so the brackets approach $c_n,$ and the whole expression behaves like ${c_nx^n}.$ For example, ${p(x)=x^3-6x^2+9x+1}$ and $x^3$ are very different near the origin, but from far away their graphs almost coincide:

The difference ${p(x)-x^3}={-6x^2+9x+1}$ does not approach $0\text{:}$ the two graphs do not come close to each other. The difference is only small compared with $x^3,$ and that is enough for the two functions to have the same limits at $+\infty$ and at $-\infty.$

Example 7. ${\lim_{x\to-\infty}(x^3-6x^2+9x+1)}={\lim_{x\to-\infty}x^3}={-\infty}$   and   ${\lim_{x\to+\infty}(1+5x^3-2x^4)}={\lim_{x\to+\infty}(-2x^4)}={{-\infty}.}$

Rational functions at infinity

The same idea, applied to the numerator and to the denominator, gives a quick method for rational functions. If ${c_n\ne0}$ and ${d_m\ne0},$ then

${\lim_{x\to\pm\infty}\frac{c_0+c_1x+\ldots+c_nx^n}{d_0+d_1x+\ldots+d_mx^m}}={{\lim_{x\to\pm\infty}\frac{c_nx^n}{d_mx^m}}.}$

So only the leading terms matter:

• if ${n\lt m}$ (the degree of the numerator is smaller), the limit is $0;$

• if ${n=m},$ the limit is ${\frac{c_n}{d_m}},$ the quotient of the leading coefficients;

• if ${n\gt m},$ the limit is $+\infty$ or $-\infty;$ the sign of ${\frac{c_nx^n}{d_mx^m}}$ decides which.

Example 8.

a) ${\lim_{x\to+\infty}\frac{3x^2-x+1}{2x^2+5}}={\lim_{x\to+\infty}\frac{3x^2}{2x^2}}={{\frac{3}{2}}.}$

The same result is obtained by dividing the numerator and the denominator by $x^2\text{:}$ ${\frac{3-\frac{1}{x}+\frac{1}{x^2}}{2+\frac{5}{x^2}}}$ approaches ${\frac{3-0+0}{2+0}}={\frac{3}{2}}.$

b) ${\lim_{x\to-\infty}\frac{4x+1}{x^2-3}}={\lim_{x\to-\infty}\frac{4x}{x^2}}={\lim_{x\to-\infty}\frac{4}{x}}={{0}.}$

c) ${\lim_{x\to-\infty}\frac{x^3-1}{2x+5}}={\lim_{x\to-\infty}\frac{x^3}{2x}}={\lim_{x\to-\infty}\frac{x^2}{2}}={{+\infty}.}$

d) ${\lim_{x\to+\infty}\frac{5-2x^3}{x+1}}={\lim_{x\to+\infty}\frac{-2x^3}{x}}={\lim_{x\to+\infty}(-2x^2)}={{-\infty}.}$

A Strategy for Finding Limits

1. Substitute ${x=a}.$ If the function is a polynomial, a rational function whose denominator is not $0$ at $a,$ or an $n$th root ${\sqrt[n]{r(x)}}$ of such a function (with ${r(a)\gt0}$ if $n$ is even), the number you get is the limit.
2. If you get ${\frac{\text{number}}{0}}$ with a number ${\ne0},$ the one-sided limits are $+\infty$ or $-\infty\text{:}$ find the sign of the fraction on each side of $a.$
3. If you get ${\frac{0}{0}},$ simplify first: factor and cancel, or multiply by the conjugate. Then substitute again.
4. For ${x\to\pm\infty},$ keep only the leading terms (or divide the numerator and the denominator by the highest power of $x$ in the denominator).

Common Mistakes

Mistake	Wrong	Right
The quotient law with denominator limit $0$	${\lim_{x\to2}\frac{x^2-4}{x-2}=\frac{0}{0}},$ so the limit does not exist	The quotient law needs a nonzero limit of the denominator. Cancel first: the limit is $4$
${\frac{0}{0}}$ as a number	${\frac{0}{0}=0}$ or ${\frac{0}{0}=1}$	${\frac{0}{0}}$ is not a number; it only says that more work is needed
Ignoring the side	${\lim_{x\to4}\frac{2-x}{(x-4)(x+2)}}={-\infty}$	From the right $-\infty,$ from the left $+\infty\text{:}$ the limit does not exist
Odd powers at $-\infty$	${\lim_{x\to-\infty}x^3=+\infty}$	${\lim_{x\to-\infty}x^3=-\infty}\text{:}$ as ${x\to-\infty},$ $x^n$ goes to $+\infty$ for even $n$ and to $-\infty$ for odd $n$
"${\infty-\infty=0}$"	${\lim_{x\to+\infty}(x^3-x^2)}={\infty-\infty}={0}$	${\lim_{x\to+\infty}(x^3-x^2)}={\lim_{x\to+\infty}x^3}={+\infty}$
Thinking that the denominator wins	${\lim_{x\to+\infty}\frac{3x^2-x+1}{2x^2+5}=0}$	Both grow; the leading terms give ${\frac{3}{2}}$

Practice Problems

1. Find ${\lim_{x\to-1}(2x^3-x+4)}.$

Answer

Substitution: ${2(-1)^3-(-1)+4}={-2+1+4}={{3}.}$

2. Find ${\lim_{x\to3}\frac{x^2+1}{x-1}}.$

Answer

The denominator is not $0$ at $3\text{:}$ ${\frac{9+1}{3-1}}={5}.$

3. Find ${\lim_{x\to-3}\frac{x^2+2x-3}{x+3}}.$

Answer

${\frac{0}{0}}.$ ${x^2+2x-3}={(x+3)(x-1)},$ so the limit is ${\lim_{x\to-3}(x-1)}={-4}.$

4. Find ${\lim_{x\to1}\frac{x^3-1}{x-1}}.$

Answer

${x^3-1}={(x-1)(x^2+x+1)},$ so the limit is ${1+1+1}={3}.$

5. Find ${\lim_{x\to0}\frac{\sqrt{x+9}-3}{x}}.$

Answer

Multiply by the conjugate ${\sqrt{x+9}+3}\text{:}$ the fraction becomes ${\frac{x}{x(\sqrt{x+9}+3)}}={{\frac{1}{\sqrt{x+9}+3}},}$ and the limit is ${\frac{1}{3+3}}={\frac{1}{6}}.$

6. Find ${\lim_{x\to3^-}\frac{x+1}{x-3}}$ and ${\lim_{x\to3^+}\frac{x+1}{x-3}}.$

Answer

The numerator approaches $4.$ For ${x\lt3}$ the denominator is a small negative number: $-\infty.$ For ${x\gt3}$ it is a small positive number: $+\infty.$

7. Find ${\lim_{x\to2}\frac{x+1}{(x-2)^2}}.$

Answer

The numerator approaches $3$ and the denominator is a small positive number on both sides: $+\infty.$

8. Find ${\lim_{x\to+\infty}\frac{2x^3-x+7}{5x^3+4x^2}}.$

Answer

Equal degrees: ${\frac{2}{5}}.$

9. Find ${\lim_{x\to+\infty}\frac{6x-1}{2x^2+3}}.$

Answer

The degree of the numerator is smaller: $0.$

10. Find ${\lim_{x\to-\infty}\frac{x^2+1}{x-4}}$ and ${\lim_{x\to-\infty}(1+x^2-4x^5)}.$

Answer

${\lim_{x\to-\infty}\frac{x^2}{x}}={\lim_{x\to-\infty}x}={-\infty}.$ ${\lim_{x\to-\infty}(-4x^5)}={+\infty},$ because $x^5$ is negative for ${x\lt0}.$

See also: Limits (Intuitive Introduction), Limits of sequences and the asymptotes calculator.