Limits Of A Rational Function

Main Article: Epsilon-Delta Definition of a Limit

The precise definition of the limit is discussed in the wiki Epsilon-Delta Definition of a Limit.

Formal Definition of a Role Limit:

The limit of $f(10)$ as $x$ approaches $x_0$ is $L$ , i.eastward.

$\lim _{ x \to x_{0} }{f(ten) } = 50$

if, for every $\epsilon > 0$ , in that location exists $\delta >0$ such that, for all $x$ ,

$0 < \left| x - x_{0} \right |<\delta \textrm{ } \implies \textrm{ } \left |f(x) - L \right| < \epsilon.$

In practice, this definition is only used in relatively unusual situations. For many applications, information technology is easier to use the definition to testify some basic properties of limits and to utilise those properties to respond straightforward questions involving limits.

The well-nigh of import properties of limits are the algebraic backdrop, which say substantially that limits respect algebraic operations:

Suppose that $\lim\limits_{x\to a} f(x) = K$ and $\lim\limits_{ten\to a} g(ten) = N.$ Then

$\begin{aligned} \lim\limits_{x\to a} \big(f(ten)+g(x)\big) &= M+N \\\\ \lim\limits_{x\to a} \big(f(x)-thou(10)\big) &= K-N \\\\ \lim\limits_{10\to a} \big(f(ten)chiliad(x)\big) &= MN \\\\ \lim\limits_{x\to a} \left(\frac{f(ten)}{g(10)}\correct) &= \frac MN \ \ \text{ (if } N\ne 0) \\\\ \lim\limits_{10\to a} f(x)^one thousand &= Chiliad^chiliad \ \ \text{ (if } Chiliad,k > 0). \end{aligned}$

These tin all be proved via application of the epsilon-delta definition. Annotation that the results are only true if the limits of the private functions be: if $\lim\limits_{x\to a} f(x)$ and $\lim\limits_{ten\to a} thou(x)$ do not be, the limit of their sum (or difference, product, or quotient) might nevertheless be.

Coupled with the basic limits $\lim_{x\to a} c = c,$ where $c$ is a constant, and $\lim_{10\to a} ten = a,$ the properties can be used to deduce limits involving rational functions:

Let $f(x)$ and $g(10)$ be polynomials, and suppose $g(a) \ne 0.$ Then

$\lim_{x\to a} \frac{f(ten)}{g(x)} = \frac{f(a)}{g(a)}.$

This is an case of continuity, or what is sometimes called limits past exchange.

Note that $g(a)=0$ is a more than difficult case; see the Indeterminate Forms wiki for further discussion.

Let $g$ and $north$ be positive integers. Notice

$\lim_{x\to i} \frac{10^one thousand-1}{x^n-1}.$

Immediately substituting $x=ane$ does not work, since the denominator evaluates to $0.$ First, separate top and bottom by $ten-ane$ to go

$\frac{x^{m-1}+ten^{m-two}+\cdots+i}{x^{due north-one}+ten^{n-2}+\cdots+1}.$

Plugging in $x=i$ to the denominator does non requite $0,$ so the limit is this fraction evaluated at $ten=one,$ which is

$\frac{1^{m-1}+1^{thousand-2}+\cdots+1}{ane^{n-1}+1^{n-2}+\cdots+ane} = \frac{grand}{n}.\ _\square$

It is important to detect that the manipulations in the in a higher place example are justified by the fact that $\lim\limits_{x\to a} f(x)$ is independent of the value of $f(x)$ at $10=a,$ or whether that value exists. This justifies, for example, dividing the top and bottom of the fraction $\frac{x^one thousand-ane}{ten^due north-one}$ by $x-1,$ since this is nonzero for $ten\ne i.$

$\lim _{x\rightarrow 10} \frac{x^{3}-10x^{2}-25x+250}{x^{four}-149x^{2}+4900} = \frac{a}{b},$

where $a$ and $b$ are coprime integers, what is $a+b?$

A one-sided limit only considers values of a function that approaches a value from either above or below.

The right-side limit of a role $f$ as it approaches $a$ is the limit

$\lim_{x \to a^+} f(ten) = L.$

The left-side limit of a function $f$ is

$\lim_{x \to a^-} f(ten) = L.$

The notation " $x \to a^-$ " indicates that we only consider values of $x$ that are less than $a$ when evaluating the limit. Likewise, for " $10 \to a^+,$ " nosotros consider only values greater than $a$ . One-sided limits are of import when evaluating limits containing absolute values $|ten|$ , sign $\text{sgn}(ten)$ , floor functions $\lfloor 10 \rfloor$ , and other piecewise functions.

The prototype above demonstrates both left- and correct-sided limits on a continuous office $f(x).$

Find the left- and correct-side limits of the signum function $\text{sgn}(x)$ as $10 \to 0:$

$\text{sgn}(x)= \begin{cases} \frac{|x|}{x} && 10\neq 0 \\ 0 && x = 0. \cease{cases}$

Consider the following graph:

From this we come across $\displaystyle \lim_{x \to 0^+} \text{sgn}(x) = 1$ and $\displaystyle \lim_{x \to 0^-}\text{sgn}(x) = -i.\ _\square$

Determine the limit $\lim\limits_{ten \to 1^{-}} \frac{\sqrt{2x}(ten-1)}{|10-1|}.$

Note that, for $x<1,$ $\left | x-one\right |$ tin can be written as $-(x-ane)$ . Hence, the limit is $\lim\limits_{x \to 1^{-}} \frac{\sqrt{2x}(x-1)}{-(x-1)} = -\sqrt{ii}.\ _\square$

By definition, a 2-sided limit

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

exists if the ane-sided limits $\displaystyle \lim_{x \to a^+} f(x)$ and $\displaystyle \lim_{x \to a^-} f(10)$ are the same.

Compute the limit

$\lim_{x \to i} \frac{|x - 1|}{x - 1} .$

Since the accented value role $f(x) = |x|$ is defined in a piecewise way, we have to consider ii limits: $\lim\limits_{10 \to 1^+} \frac{|ten - ane|}{10 - 1}$ and $\lim\limits_{x \to ane^-} \frac{|ten - ane|}{x - 1}.$

Starting time with the limit $\lim\limits_{x \to 1^+} \frac{|x - ane|}{ten - one}.$ For $x>ane,$ $|x - 1| = x -1.$ So

$\lim_{x \to 1^+} \frac{|x - i|}{x - 1} =\lim_{x \to 1^+} \frac{ten - 1}{ten - i} =1.$

Let united states now consider the left-mitt limit

$\lim_{x \to 1^-} \frac{|x - 1|}{x - ane}.$

For $ten<i,$ $x - 1 = -|x-1|.$ So

$\lim_{x \to 1^-} \frac{|x-ane|}{-|x - 1|} = -i .$

So the two-sided limit $\lim\limits_{x \to 1} \frac{|x - 1|}{x - 1}$ does not exist. $_\foursquare$

The paradigm below is a graph of a function $f(x)$ . As shown, it is continuous for all points except $x = -one$ and $10=ii$ which are its asymptotes. Find all the integer points $-4 <I < 4,$ where the two-sided limit $\lim_{x \to I} f(x)$ exists.

image

Since the graph is continuous at all points except $x=-1$ and $x=two$ , the two-sided limit exists at $x=-3$ , $x=-2$ , $x=0$ , $ten=one,$ and $x=three$ . At $x=2,$ there is no finite value for either of the two-sided limits, since the office increases without bound equally the $ten$ -coordinate approaches $2$ (merely meet the side by side section for a further discussion). The situation is similar for $10=-1.$ So the points $10=-three$ , $x=-two$ , $ten=0$ , $x=ane,$ and $x=iii$ are all the integers on which two-sided limits are defined. $_\square$

As seen in the previous department, one manner for a limit non to be is for the one-sided limits to disagree. Another common way for a limit to not exist at a betoken $a$ is for the function to "blow up" near $a,$ i.due east. the role increases without jump. This happens in the above example at $ten=ii,$ where in that location is a vertical asymptote. This mutual state of affairs gives rise to the post-obit notation:

Given a office $f(x)$ and a real number $a,$ we say

$\lim_{10\to a} f(x) = \infty.$

If the function tin can be made arbitrarily large by moving $x$ sufficiently close to $a,$

$\text{for all } N>0, \text{ in that location exists } \delta>0 \text{ such that } 0<|x-a|<\delta \implies f(x)>Northward.$

There are similar definitions for 1-sided limits, as well as limits "approaching $-\infty$ ."

Alert: If $\lim\limits_{x\to a} f(x) = \infty,$ it is tempting to say that the limit at $a$ exists and equals $\infty.$ This is incorrect. If $\lim\limits_{x\to a} f(10) = \infty,$ the limit does not exist; the note just gives data about the way in which the limit fails to exist, i.e. the value of the function "approaches $\infty$ " or increases without spring as $x \rightarrow a$ .

What tin we say most $\lim\limits_{x \to 0} \frac{1}{x}?$

Separating the limit into $\lim\limits_{10 \to 0^+} \frac{i}{x}$ and $\lim\limits_{x \to 0^-} \frac{1}{x}$ , we obtain

$\lim_{10 \to 0^+} \frac{1}{x} = \infty$

and

$\lim_{ten \to 0^-} \frac{i}{x} = -\infty.$

To prove the first argument, for any $North>0$ in the formal definition, nosotros can take $\delta = \frac1N,$ and the proof of the second statement is like.

So the function increases without leap on the right side and decreases without bound on the left side. We cannot say anything else about the two-sided limit $\lim\limits_{10\to a} \frac1{10} \ne \infty$ or $-\infty.$ Contrast this with the adjacent example. $_\square$

What can we say most $\lim\limits_{ten \to 0} \frac{ane}{x^2}?$

Separating the limit into $\lim\limits_{x \to 0^+} \frac{ane}{10^2}$ and $\lim\limits_{ten \to 0^-} \frac{one}{10^2}$ , we obtain

$\lim_{x \to 0^+} \frac{one}{10^2} = \infty$

and

$\lim_{x \to 0^-} \frac{1}{x^2} = \infty.$

Since these limits are the same, we have $\lim_{x \to 0} \frac{1}{x^2} = \infty .$ Again, this limit does not, strictly speaking, exist, but the statement is meaningful all the same, as it gives data about the beliefs of the function $\frac1{x^2}$ near $0.$ $_\square$

Let

$f(x)=\frac{a_0 x^{yard}+a_1 10^{m+1}+\cdots +a_k x^{thou+chiliad}}{b_0 x^{due north}+b_1 x^{n+1}+\cdots +b_ 50 ten^{northward+l}},$

where $a_0 \neq 0, b_0 \neq 0,$ and $one thousand,due north \in \mathbb N.$

Then given (A), (B), (C), or (D), $\displaystyle\lim_{ten\rightarrow 0}f(x)$ equals which of (i), (2), (three), and (4)?

Match the columns:

Cavalcade-I	Cavalcade-II
(A) if $m>n$	(1) $\infty$
(B) if $g=n$	(2) $-\infty$
(C) if $chiliad<north,$ $due north-thousand$ is even, and $\frac{a_0}{b_0}>0$ $\hspace{10mm}$	(iii) $\frac{a_0}{b_0}$
(D) if $m<northward,$ $north-grand$ is even, and $\frac{a_0}{b_0}<0$ $\hspace{10mm}$	(4) $0$

Notation: For case, if (A) correctly matches (ane), (B) with (two), (C) with (iii), and (D) with (4), then answer as 1234.

Some other extension of the limit concept comes from considering the function's behavior equally $x$ "approaches $\infty$ ," that is, as $ten$ increases without bound.

The equation $\lim\limits_{x\to\infty} f(10) = L$ means that the values of $f$ tin exist made arbitrarily close to $L$ by taking $x$ sufficiently large. That is,

$\text{for all } \epsilon > 0, \text{ at that place is } North>0 \text{ such that } 10>N \implies |f(10)-L|<\epsilon.$

At that place are similar definitions for $\lim\limits_{x\to -\infty} f(x) = L,$ also as $\lim\limits_{x\to\infty} f(x) = \infty,$ and then on.

Graphically, $\lim\limits_{x\to a} f(x) = \infty$ corresponds to a vertical asymptote at $a,$ while $\lim\limits_{ten\to\infty} f(10) = Fifty$ corresponds to a horizontal asymptote at $L.$

Main Commodity: Limits past Factoring

Limits by factoring refers to a technique for evaluating limits that requires finding and eliminating common factors.

Primary Commodity: Limits by Commutation

Evaluating limits past substitution refers to the idea that under certain circumstances (namely if the function we are examining is continuous), we can evaluate the limit by simply evaluating the part at the point nosotros are interested in.

Master Commodity: 50'Hôpital's Rule

L'Hôpital's rule is an approach to evaluating limits of certain quotients by means of derivatives. Specifically, under certain circumstances, it allows united states to replace $\lim \frac{f(x)}{1000(x)}$ with $\lim \frac{f'(10)}{g'(x)},$ which is frequently easier to evaluate.

Evaluate $\lim\limits_{x\to\infty} \frac{x^2 + 2x +4}{3x^two+ 4x+125345}$ .

We have

$\brainstorm{aligned} &&\displaystyle \lim_{x\to\infty} \frac{x^2 + 2x +4}{3x^2+ 4x+125345} &=& \displaystyle \lim_{x\to\infty} \frac{i + \frac2x + \frac4{ten^2}}{3+ \frac4x+ \frac{125345}{x^ii}} &=& \displaystyle \frac{1+0+0}{3+0+0} = \frac13.\ _\square \end{aligned}$