At what point does both right hand and left hand limits exist, but the limit does not exist? Give your...












-1












$begingroup$


Question 1



Consider the function defined by
$$f(x) =
begin{cases}
x^2 - 1, & text{if $x leq 0$}\
x - 2, & text{if $0 < x < 1$}\
c, & text{if $x = 1$}\
-x, & text{if $x > 1$}
end{cases}
$$



(a) At what point does both right hand and left hand limits exist, but the limit does not exist? Give your reason.
(4 marks)



(b) Find the value of $c$ for which $f(x)$ is continuous at $x=1$. Give your reason.
(6 marks)



Current ANSWER:
enter image description here
not sure if answer a is correct.
not sure about the method to solve question b










share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to MathSE. When you pose a question on this site, it is expected that you include your own thoughts on the problem. Please edit the question to tell us what you know, show what you have attempted, and explain where you are stuck so that you receive answers that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Dec 8 '18 at 11:16






  • 1




    $begingroup$
    Also, I do not understand how the title relates to the question. Please revise your title.
    $endgroup$
    – N. F. Taussig
    Dec 8 '18 at 11:23










  • $begingroup$
    Your working for part (a) seems very clear. What in particular is your reason for doubt?
    $endgroup$
    – David K
    Dec 8 '18 at 14:15
















-1












$begingroup$


Question 1



Consider the function defined by
$$f(x) =
begin{cases}
x^2 - 1, & text{if $x leq 0$}\
x - 2, & text{if $0 < x < 1$}\
c, & text{if $x = 1$}\
-x, & text{if $x > 1$}
end{cases}
$$



(a) At what point does both right hand and left hand limits exist, but the limit does not exist? Give your reason.
(4 marks)



(b) Find the value of $c$ for which $f(x)$ is continuous at $x=1$. Give your reason.
(6 marks)



Current ANSWER:
enter image description here
not sure if answer a is correct.
not sure about the method to solve question b










share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to MathSE. When you pose a question on this site, it is expected that you include your own thoughts on the problem. Please edit the question to tell us what you know, show what you have attempted, and explain where you are stuck so that you receive answers that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Dec 8 '18 at 11:16






  • 1




    $begingroup$
    Also, I do not understand how the title relates to the question. Please revise your title.
    $endgroup$
    – N. F. Taussig
    Dec 8 '18 at 11:23










  • $begingroup$
    Your working for part (a) seems very clear. What in particular is your reason for doubt?
    $endgroup$
    – David K
    Dec 8 '18 at 14:15














-1












-1








-1





$begingroup$


Question 1



Consider the function defined by
$$f(x) =
begin{cases}
x^2 - 1, & text{if $x leq 0$}\
x - 2, & text{if $0 < x < 1$}\
c, & text{if $x = 1$}\
-x, & text{if $x > 1$}
end{cases}
$$



(a) At what point does both right hand and left hand limits exist, but the limit does not exist? Give your reason.
(4 marks)



(b) Find the value of $c$ for which $f(x)$ is continuous at $x=1$. Give your reason.
(6 marks)



Current ANSWER:
enter image description here
not sure if answer a is correct.
not sure about the method to solve question b










share|cite|improve this question











$endgroup$




Question 1



Consider the function defined by
$$f(x) =
begin{cases}
x^2 - 1, & text{if $x leq 0$}\
x - 2, & text{if $0 < x < 1$}\
c, & text{if $x = 1$}\
-x, & text{if $x > 1$}
end{cases}
$$



(a) At what point does both right hand and left hand limits exist, but the limit does not exist? Give your reason.
(4 marks)



(b) Find the value of $c$ for which $f(x)$ is continuous at $x=1$. Give your reason.
(6 marks)



Current ANSWER:
enter image description here
not sure if answer a is correct.
not sure about the method to solve question b







functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 11:51







auxy12

















asked Dec 8 '18 at 11:01









auxy12auxy12

257




257












  • $begingroup$
    Welcome to MathSE. When you pose a question on this site, it is expected that you include your own thoughts on the problem. Please edit the question to tell us what you know, show what you have attempted, and explain where you are stuck so that you receive answers that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Dec 8 '18 at 11:16






  • 1




    $begingroup$
    Also, I do not understand how the title relates to the question. Please revise your title.
    $endgroup$
    – N. F. Taussig
    Dec 8 '18 at 11:23










  • $begingroup$
    Your working for part (a) seems very clear. What in particular is your reason for doubt?
    $endgroup$
    – David K
    Dec 8 '18 at 14:15


















  • $begingroup$
    Welcome to MathSE. When you pose a question on this site, it is expected that you include your own thoughts on the problem. Please edit the question to tell us what you know, show what you have attempted, and explain where you are stuck so that you receive answers that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Dec 8 '18 at 11:16






  • 1




    $begingroup$
    Also, I do not understand how the title relates to the question. Please revise your title.
    $endgroup$
    – N. F. Taussig
    Dec 8 '18 at 11:23










  • $begingroup$
    Your working for part (a) seems very clear. What in particular is your reason for doubt?
    $endgroup$
    – David K
    Dec 8 '18 at 14:15
















$begingroup$
Welcome to MathSE. When you pose a question on this site, it is expected that you include your own thoughts on the problem. Please edit the question to tell us what you know, show what you have attempted, and explain where you are stuck so that you receive answers that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Dec 8 '18 at 11:16




$begingroup$
Welcome to MathSE. When you pose a question on this site, it is expected that you include your own thoughts on the problem. Please edit the question to tell us what you know, show what you have attempted, and explain where you are stuck so that you receive answers that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Dec 8 '18 at 11:16




1




1




$begingroup$
Also, I do not understand how the title relates to the question. Please revise your title.
$endgroup$
– N. F. Taussig
Dec 8 '18 at 11:23




$begingroup$
Also, I do not understand how the title relates to the question. Please revise your title.
$endgroup$
– N. F. Taussig
Dec 8 '18 at 11:23












$begingroup$
Your working for part (a) seems very clear. What in particular is your reason for doubt?
$endgroup$
– David K
Dec 8 '18 at 14:15




$begingroup$
Your working for part (a) seems very clear. What in particular is your reason for doubt?
$endgroup$
– David K
Dec 8 '18 at 14:15










1 Answer
1






active

oldest

votes


















1












$begingroup$

b) for continuity at x=1 at first we have to do L.H.L $f(x)$ at 1 = R.H.L $f(x)$ at 1 =$lim$ x tends to 1 $f(x)$ .
Here ,
L.H.L. of $f(x)$ at x=1 = R.H.L. of $f(x)$ at x=1 = -1
So, for continuity c must be -1
Sorry,for this type of writing.I am not too much familiar with this.






share|cite|improve this answer









$endgroup$













    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
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030960%2fat-what-point-does-both-right-hand-and-left-hand-limits-exist-but-the-limit-doe%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









    1












    $begingroup$

    b) for continuity at x=1 at first we have to do L.H.L $f(x)$ at 1 = R.H.L $f(x)$ at 1 =$lim$ x tends to 1 $f(x)$ .
    Here ,
    L.H.L. of $f(x)$ at x=1 = R.H.L. of $f(x)$ at x=1 = -1
    So, for continuity c must be -1
    Sorry,for this type of writing.I am not too much familiar with this.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      b) for continuity at x=1 at first we have to do L.H.L $f(x)$ at 1 = R.H.L $f(x)$ at 1 =$lim$ x tends to 1 $f(x)$ .
      Here ,
      L.H.L. of $f(x)$ at x=1 = R.H.L. of $f(x)$ at x=1 = -1
      So, for continuity c must be -1
      Sorry,for this type of writing.I am not too much familiar with this.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        b) for continuity at x=1 at first we have to do L.H.L $f(x)$ at 1 = R.H.L $f(x)$ at 1 =$lim$ x tends to 1 $f(x)$ .
        Here ,
        L.H.L. of $f(x)$ at x=1 = R.H.L. of $f(x)$ at x=1 = -1
        So, for continuity c must be -1
        Sorry,for this type of writing.I am not too much familiar with this.






        share|cite|improve this answer









        $endgroup$



        b) for continuity at x=1 at first we have to do L.H.L $f(x)$ at 1 = R.H.L $f(x)$ at 1 =$lim$ x tends to 1 $f(x)$ .
        Here ,
        L.H.L. of $f(x)$ at x=1 = R.H.L. of $f(x)$ at x=1 = -1
        So, for continuity c must be -1
        Sorry,for this type of writing.I am not too much familiar with this.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 13:37









        Supriyo BanerjeeSupriyo Banerjee

        1176




        1176






























            draft saved

            draft discarded




















































            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030960%2fat-what-point-does-both-right-hand-and-left-hand-limits-exist-but-the-limit-doe%23new-answer', 'question_page');
            }
            );

            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







            Popular posts from this blog

            Plaza Victoria

            Puebla de Zaragoza

            Musa