Tutorial: Limits: algebraic viewpoint

Adaptive game version

This tutorial: Part A: Calculating limits algebraically

Go to Part B: Calculating limits at infinity

(This topic is also in Section 10.2 in Finite Mathematics and Applied Calculus)

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%%Note You need to be familiar with the notion of a limit and how to interpret them numerically and geometrically before starting work on this tutorial. If you feel you need to review that material, go back to the %%numericallimitstut.

Limits and closed form functions

#[Consider the following limit:][Considera el siguiente límite:]#

$\displaystyle \lim_{x \to 2} \frac{x^2 - 3x}{4x + 3}.$

If you estimate this limit either %%numerically or %%geometrically, you will find that

$\displaystyle \lim_{x \to 2} \frac{x^2 - 3x}{4x + 3} \approx -0.1818.$

But, notice that you can obtain this answer, or an even more precise answer, by simply substituting $x = 2$ in the given function:

$f(x) = \dfrac{x^2 - 3x}{4x + 3}$ \\ $f(2) = \dfrac{4 - 6}{8 + 3} = -\dfrac{2}{11} = -0.181818...$

Not only is this answer more accurate than the one coming from numerical or graphical method but in fact it gives the exact limit.

#[Q][P]#: Is that all there is to evaluating limits algebraically: just substitute the number that $x$ is approaching in the given expression?
#[A][R]#: We could do that in the case above for two reasons:

The function whose limit we are taking is specified by a single "nice" formula, and
The number that $x$ is approaching is in the domain of the function: substituting it results in a real number.

Let's make this idea more precise:

Closed form functions

A function $f$ is written in closed form if $f(x)$ specified by combining constants, powers of $x$, exponential functions, radicals, logarithms, absolute values, trigonometric functions (and certain other functions) into a single mathematical formula by means of the usual arithmetic operations and composition of functions. A closed-form function is any function that can be written in closed form.

Examples

#[The functions][Las funciones]# $f(x) = 2x^2 - 7x + \dfrac{3 - x^2}{x}$, $g(x) = \dfrac{\sqrt{x^2 - 1}}{6x + 4}$, #[and][y]# $h(x) = e^{-4x^2-2}$ #[are all closed-form functions.][son todas funciones de forma cerrada.]#
#[The function][La función]#

#[is not closed form because $f(x)$ is not expressed by a single mathematical formula, and nor can it be expressed in such a way.][no es una forma cerrada porque $f(x)$ no se expresa mediante una fórmula matemática única, y tampoco se puede expresar de esa manera.]#

#[Theorem C: Limits of closed form functions][Teorema C: Límites de funciones de forma cerrada]#

#[If $f$ is a closed-form function and $f(a)$ is defined, then $\displaystyle \lim_{x \to a} f(x)$ exists, and equals $f(a).$][Si $f$ es una función de forma cerrada y $f(a)$ está definida, entonces $\displaystyle \lim_{x \to a} f(x)$ existe y es igual a $f(a).$]#

Examples

$f(x) = \dfrac{2x^2+7x+3}{x + 3}$ is a closed-form function. and $2$ is in the domain of $f$. So,
$\displaystyle \lim_{x \to 2} \frac{2x^2+7x+3}{x+3} = f(2) = \frac{25}{5} = 5$.
Because $-1$ is also in the domain of $f$,
$\displaystyle \lim_{x \to -1} \frac{2x^2+7x+3}{x+3} = f(-1) -\frac{-2}{2} = -1$.
However, $\displaystyle \lim_{x \to -3} \frac{2x^2+7x+3}{x+3}$ cannot be obtained by substitution, as $-3$ is not in the domain of $f$.

Some for you

Limits at singular points

Let's take another look at one of the limits above that we could not calculate by substitution::

$\displaystyle \lim_{x \to -3} \frac{2x^2+7x+3}{x+3}$.

Substitution does not work because $-3$ is not in the domain $f(x) = \dfrac{x^2 - 3x}{x + 3}$. Put another way, substituting $x = -3$ results in $0/0$, which does not make sense as a real number: the function $f$ is singular at $-3$: This means that $f$ is undefined at $-3$ although it is defined for values nearby (we looked at singular points geometrically in %%geometriclimitstut).:

#[Q][P]#: So, how do we calculate limits like that?
#[A][R]#: #[One way that often works is to remove the singularity by modifying the function by cancellation so as to result in a new closed-form function that is no longer singular at that point:][Una forma que suele funcionar es eliminar la singularidad modificando la función por cancelación para que resulte en una nueva función de forma cerrada que ya no es singular en ese punto:]#

$\dfrac{2x^2+7x+3}{x+3}$ \t ${}= \dfrac{(2x+1)\color{blue}{(x+3)}}{\color{blue}{x+3}}$ \t #[Singular at][Singular en]# $x=-3$ \\ \t ${}= 2x+1 \quad (x \ne -3)$ \t #[Not singular at][Nosingular en]# $x=-3$

#[The singularity is gone! We have sneakily changed the function to a new one that is no longer singular. When $x \ne -3$ these two functions are the same, so their limits as $x \to -3$ agree; mathematically, the limit depends only on what is happening near $x = -3$. Thus,][¡La singularidad se ha ido! Hemos cambiado furtivamente la función a una nueva que ya no es singular. Cuando $x \ne -3$ estas dos funciones son iguales, entonces sus límites como $x \to -3$ concuerdan; matemáticamente, el límite depende solo de lo que sucede cerca de $x = -3$. De este modo,]#

$\displaystyle \lim_{x \to -3}\dfrac{2x^2+7x+3}{x+3}$ \t ${}= \lim_{x \to -3}\dfrac{(2x+1)\color{blue}{(x+3)}}{\color{blue}{x+3}}$ \\ \t $\displaystyle {}= \lim_{x \to -3}2x+1$ \\ \t ${}= 2(-3) + 1 = -5$ \t $2x+1$ #[is closed form with $-3$ in its domain][es de forma cerrada con $-3$ en su dominio]#

Determinate and indeterminate forms

Notice something about the limits we calculated above by canceling to remove the singularities: In each of those, if we had just subtituted for $x$ at the singular point without canceling we would have obtained $0/0$ as both the numerator and denominator have the value $0$ at the singular point. However, in each case we got a different limit. Also, there are instances like this where the limit does not exist or is infinite:

$\displaystyle \lim_{x \to -3} \frac{2x^2+7x+3}{x+3} = -5 \quad$ \t #[The example we did above][El ejemplo hicimos arriba]# \\ #[Substituting][Sustituir]# $x = -3$ #[results in][resulta en]# $\dfrac{0}{0}$. \\ $\displaystyle \lim_{x \to 0} \frac{x}{x^3} = \infty$ \t #[Cancelling an $x$ gives $\dfrac{1}{x^2}$ which gets large positive without bound a $x \to 0$.][Cancelar un $x$ da $\dfrac{1}{x^2}$ que se vuelve positivo grande sin límite a $x \to 0$.]#. \\ #[Substituting][Sustituir]# $x = 0$ #[results in][resulta en]# $\dfrac{0}{0}$. \\ $\displaystyle \lim_{x \to 0} \frac{|x|}{x}$ #[does not exist.][no existe.]# \t #[For positive $x$ the fraction is $1$, and for negative $x$ it is $-1$.][Para $x$ positivo la fracción es $1$, y para $x$ negativo es $-1$.]#. \\ #[Substituting][Sustituir]# $x = 0$ #[results in][resulta en]# $\dfrac{0}{0}$.

#[These examples suggest that the expression $0/0$ tells us nothing at all about the behavior of the limit, and for this reason we call $0/0$ an indeterminate form.][Estos ejemplos sugieron que la expresión $0/0$ no nos dice nada sobre el comportamiento del límite, y por esta razón llamamos a $0/0$ una forma indeterminada.]#

#[If substituting $x = a$ yields the indeterminate form $0/0$, then you know absolutely nothing about the limit—even whether it exists or not. To determine what is going on, you need to simplify the function or do some kind of further analysis.][Si la sustitución $x = a$ te da la forma indeterminada 0/0, sabes nada en absoluto acerca del límite—ni siquiera sí o no existe. Para determinar lo que está pasando, debes simplificar la función o hacer cualquier tipo de análisis adicional.]#

#[Q][P]# #[What if we end up with something else, like 3/0?][¿Y si nos queda otra cosa, como 3/0, qué?]#
#[A][R]# #[Think about what happens when you divide a nonzero number (such as −3) by a number very close to zero (such as 0.000001). The result is number with very large absolute value (such as $\frac{-3}{0.000001} = -3,000,000$). It is for this reason that expression $k/0$ (where $k$ is nonzero) is referred to as a determinate form, as it always gives a definite answer for the limit:][Piensa en lo que resulta cuando dividas un número distinto de cero (como −3) por un número muy cercano a cero (como 0.000001). El resultado es entonces un número con valor absoluto muy grande (como $\frac{-3}{0.000001} = -3,000,000$). Es por eso que referimos a la expresión $k/0$ (en la que $k$ es distinto de cero) como una forma determinada, ya que nos siempre de una respuesta definida para el límite:]#

The determinate form k/0

If the substitution $x = a$ in the function $f$ yields $k/0$ with $k \ne 0$, then

a. If $f(x)$ is positive as $x \to a$, then $\displaystyle \lim_{x \to a} f(x) = \infty$.
b. If $f(x)$ is negative as $x \to a$, then $\displaystyle \lim_{x \to a} f(x) = -\infty$.
c. If $f(x)$ has both positive and negative values as $x \to a$, then $\displaystyle \lim_{x \to a} f(x)$ does not exist.

Examples

approaching 1 ↓
1. $\displaystyle \lim_{x \to 0} \frac{1}{x^2} = \infty$	$\color{blue}{\dfrac{1}{0^+}=\infty}$
↑ approaching 0⁺
Dividing 1 by a positive number very close to 0 yields a large positive number.

approaching 4 ↓
2. $\displaystyle \lim_{x \to 2} \frac{3x - 2}{(x - 2)^2} = \infty$	$\color{blue}{\dfrac{4}{0^+}=\infty}$
↑ approaching 0⁺
Dividing 4 by a positive number very close to 0 yields a large positive number.

approaching 4 ↓
3. $\displaystyle \lim_{x \to 2^-} \frac{3x - 2}{x - 2} = -\infty$	$\color{blue}{\dfrac{4}{0^-}=-\infty}$
↑ approaching 0⁻
Dividing 4 by a negative number very close to 0 yields a large negative number.

approaching 4 ↓
4. $\displaystyle \lim_{x \to 2} \frac{3x - 2}{x - 2} $ does not exist.	$\color{blue}{\dfrac{4}{0}=\pm\infty}$
↑ approaching 0
Here, $\displaystyle \color{blue}{\lim_{x \to 2^{-}} \frac{3x - 2}{x - 2} = \dfrac{4}{0^-}=-\infty}$ and $\displaystyle \color{blue}{\lim_{x \to 2^{+}} \frac{3x - 2}{x - 2} = \dfrac{4}{0^+}=\infty}$

Some for you

Here is a quiz that includes all the types of limits we have discussed so far. Some are direct applications of Theorem C, some give the indeterminate form $0/0$, and others the determinate form $k/0$. It is always best to first simplify by cancellation if possible.

Now try the exercises in Section 10.2 in Finite Mathematics and Applied Calculus. or move ahead to the next tutorial by pressing "Next tutorial" on the sidebar.