Limit of rational function solving
$begingroup$
I have to solve limit of rational function, but it turns out I do mistake somewhere. Where I do wrong? Does my calculations correct?
$$lim_{xto infty}frac{x^3-2x-1}{x^5-2x-1}$$
Step 1:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)}$$
Step 2:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)} = frac{0-0-0}{1-0-0}=frac{0}{1}=0$$
algebra-precalculus limits rational-functions
asked May 25 '17 at 10:47
TeodorKolevTeodorKolev
207212
$endgroup$
|
show 10 more comments
$begingroup$
I have to solve limit of rational function, but it turns out I do mistake somewhere. Where I do wrong? Does my calculations correct?
$$lim_{xto infty}frac{x^3-2x-1}{x^5-2x-1}$$
Step 1:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)}$$
Step 2:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)} = frac{0-0-0}{1-0-0}=frac{0}{1}=0$$
algebra-precalculus limits rational-functions
asked May 25 '17 at 10:47
TeodorKolevTeodorKolev
207212
$endgroup$
$begingroup$
So $1=1*infty$? I never knew that!
$endgroup$
– Lord Shark the Unknown
May 25 '17 at 10:49
$begingroup$
I cannot follow your every single step!!
$endgroup$
– velut luna
May 25 '17 at 10:52
$begingroup$
Please check edited answer. Is that correct?
$endgroup$
– TeodorKolev
May 25 '17 at 10:56
1
$begingroup$
The editing answer is exactly correct
$endgroup$
– Noor Aslam
May 25 '17 at 11:02
2
$begingroup$
The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
$endgroup$
– B. Mehta
May 25 '17 at 11:23
|
show 10 more comments
$begingroup$
I have to solve limit of rational function, but it turns out I do mistake somewhere. Where I do wrong? Does my calculations correct?
$$lim_{xto infty}frac{x^3-2x-1}{x^5-2x-1}$$
Step 1:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)}$$
Step 2:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)} = frac{0-0-0}{1-0-0}=frac{0}{1}=0$$
algebra-precalculus limits rational-functions
asked May 25 '17 at 10:47
TeodorKolevTeodorKolev
207212
$endgroup$
I have to solve limit of rational function, but it turns out I do mistake somewhere. Where I do wrong? Does my calculations correct?
$$lim_{xto infty}frac{x^3-2x-1}{x^5-2x-1}$$
Step 1:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)}$$
Step 2:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)} = frac{0-0-0}{1-0-0}=frac{0}{1}=0$$
algebra-precalculus limits rational-functions
algebra-precalculus limits rational-functions
asked May 25 '17 at 10:47
TeodorKolevTeodorKolev
207212
asked May 25 '17 at 10:47
TeodorKolevTeodorKolev
207212
edited May 25 '17 at 11:30
TeodorKolev
asked May 25 '17 at 10:47
TeodorKolevTeodorKolev
207212
asked May 25 '17 at 10:47
TeodorKolevTeodorKolev
207212
asked May 25 '17 at 10:47
TeodorKolevTeodorKolev
207212
207212
$begingroup$
So $1=1*infty$? I never knew that!
$endgroup$
– Lord Shark the Unknown
May 25 '17 at 10:49
$begingroup$
I cannot follow your every single step!!
$endgroup$
– velut luna
May 25 '17 at 10:52
$begingroup$
Please check edited answer. Is that correct?
$endgroup$
– TeodorKolev
May 25 '17 at 10:56
1
$begingroup$
The editing answer is exactly correct
$endgroup$
– Noor Aslam
May 25 '17 at 11:02
2
$begingroup$
The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
$endgroup$
– B. Mehta
May 25 '17 at 11:23
|
show 10 more comments
$begingroup$
So $1=1*infty$? I never knew that!
$endgroup$
– Lord Shark the Unknown
May 25 '17 at 10:49
$begingroup$
I cannot follow your every single step!!
$endgroup$
– velut luna
May 25 '17 at 10:52
$begingroup$
Please check edited answer. Is that correct?
$endgroup$
– TeodorKolev
May 25 '17 at 10:56
1
$begingroup$
The editing answer is exactly correct
$endgroup$
– Noor Aslam
May 25 '17 at 11:02
2
$begingroup$
The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
$endgroup$
– B. Mehta
May 25 '17 at 11:23
$begingroup$
So $1=1*infty$? I never knew that!
$endgroup$
– Lord Shark the Unknown
May 25 '17 at 10:49
$begingroup$
So $1=1*infty$? I never knew that!
$endgroup$
– Lord Shark the Unknown
May 25 '17 at 10:49
$begingroup$
I cannot follow your every single step!!
$endgroup$
– velut luna
May 25 '17 at 10:52
$begingroup$
I cannot follow your every single step!!
$endgroup$
– velut luna
May 25 '17 at 10:52
$begingroup$
Please check edited answer. Is that correct?
$endgroup$
– TeodorKolev
May 25 '17 at 10:56
$begingroup$
Please check edited answer. Is that correct?
$endgroup$
– TeodorKolev
May 25 '17 at 10:56
1
1
$begingroup$
The editing answer is exactly correct
$endgroup$
– Noor Aslam
May 25 '17 at 11:02
$begingroup$
The editing answer is exactly correct
$endgroup$
– Noor Aslam
May 25 '17 at 11:02
2
2
$begingroup$
The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
$endgroup$
– B. Mehta
May 25 '17 at 11:23
$begingroup$
The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
$endgroup$
– B. Mehta
May 25 '17 at 11:23
|
show 10 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.
For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.
Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
{text{if};;limlimits_{x to a} gleft( x right) ne 0}$$ to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.
To Sum It Up
Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$
Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:
$bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$
$bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$
$bullet$ If $n<m$ then the limit is $0$.
$bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.
Note that this method is only for limits to infinity
answered Dec 8 '18 at 11:45
E.NoleE.Nole
166114
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2296069%2flimit-of-rational-function-solving%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.
For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.
Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
{text{if};;limlimits_{x to a} gleft( x right) ne 0}$$ to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.
To Sum It Up
Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$
Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:
$bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$
$bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$
$bullet$ If $n<m$ then the limit is $0$.
$bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.
Note that this method is only for limits to infinity
answered Dec 8 '18 at 11:45
E.NoleE.Nole
166114
$endgroup$
add a comment |
$begingroup$
The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.
For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.
Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
{text{if};;limlimits_{x to a} gleft( x right) ne 0}$$ to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.
To Sum It Up
Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$
Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:
$bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$
$bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$
$bullet$ If $n<m$ then the limit is $0$.
$bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.
Note that this method is only for limits to infinity
answered Dec 8 '18 at 11:45
E.NoleE.Nole
166114
$endgroup$
add a comment |
$begingroup$
The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.
For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.
Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
{text{if};;limlimits_{x to a} gleft( x right) ne 0}$$ to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.
To Sum It Up
Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$
Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:
$bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$
$bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$
$bullet$ If $n<m$ then the limit is $0$.
$bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.
Note that this method is only for limits to infinity
answered Dec 8 '18 at 11:45
E.NoleE.Nole
166114
$endgroup$
The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.
For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.
Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
{text{if};;limlimits_{x to a} gleft( x right) ne 0}$$ to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.
To Sum It Up
Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$
Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:
$bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$
$bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$
$bullet$ If $n<m$ then the limit is $0$.
$bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.
Note that this method is only for limits to infinity
answered Dec 8 '18 at 11:45
E.NoleE.Nole
166114
answered Dec 8 '18 at 11:45
E.NoleE.Nole
166114
answered Dec 8 '18 at 11:45
E.NoleE.Nole
166114
answered Dec 8 '18 at 11:45
E.NoleE.Nole
166114
166114
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2296069%2flimit-of-rational-function-solving%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
So $1=1*infty$? I never knew that!
$endgroup$
– Lord Shark the Unknown
May 25 '17 at 10:49
$begingroup$
I cannot follow your every single step!!
$endgroup$
– velut luna
May 25 '17 at 10:52
$begingroup$
Please check edited answer. Is that correct?
$endgroup$
– TeodorKolev
May 25 '17 at 10:56
1
$begingroup$
The editing answer is exactly correct
$endgroup$
– Noor Aslam
May 25 '17 at 11:02
2
$begingroup$
The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
$endgroup$
– B. Mehta
May 25 '17 at 11:23