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Math Help - Continuous limit

  1. #1
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    Continuous limit

    Can you tell me if I am on the right track on this problem

    If f and g are continuous functions with f(-3) = 5 and the following limit, find g(-3).

    <br />
\lim_{x \to -3} [4f(x)-g(x)] = 13<br />


    My work:

    <br />
\lim_{x \to -3}g(-3) = ?<br />

    <br />
4\lim_{x \to -3}f(-3) = 5<br />




    <br />
g(-3) = 7<br />

    I just winged this one since I don't know how to do it

    Am I even close? lol
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by silencecloak View Post
    Can you tell me if I am on the right track on this problem

    If f and g are continuous functions with f(-3) = 5 and the following limit, find g(-3).

    <br />
\lim_{x \to -3} [4f(x)-g(x)] = 13<br />


    My work:

    <br />
\lim_{x \to -3}g(-3) = ?<br />

    <br />
4\lim_{x \to -3}f(-3) = 5<br />




    <br />
g(-3) = 7<br />

    I just winged this one since I don't know how to do it

    Am I even close? lol
    i assume they are continuous on the real line

    no, you're not close

    f(-3) = 5, since f(x) is continuous, \lim_{x \to -3}f(x) = f(-3) (you should know this), thus 4 \lim_{x \to -3}f(x) = 4 \cdot f(-3) = 4 \cdot 5 = 20
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  3. #3
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    <br /> <br />
\lim_{x \to -3} [20-g(x)] = 13<br />

    <br />
20-x = 13 <br />

    <br />
x=7<br />
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by silencecloak View Post
    <br /> <br />
\lim_{x \to -3} [20-g(x)] = 13<br />

    <br />
20-x = 13 <br />

    <br />
x=7<br />
    exactly how did you get x? how do you know \lim_{x \to -3}g(x) = x?
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    exactly how did you get x?


    <br />
4\lim_{x \to -3}f(-3) - \lim_{x \to -3}g(x) = 13<br />

    =

    <br />
20 - g(x) = 13<br /> <br />
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by silencecloak View Post
    <br />
4\lim_{x \to -3}f(-3) - \lim_{x \to -3}g(x) = 13<br />

    =

    <br />
20 - g(x) = 13<br /> <br />
    don't drop the limit from the g(x), you can't do that. you must solve for it

    note that since g(x) is continuous on the reals, \lim_{x \to -3}g(x) = g(-3), i told you the same thing for f.
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    don't drop the limit from the g(x), you can't do that. you must solve for it

    note that since g(x) is continuous on the reals, \lim_{x \to -3}g(x) = g(-3), i told you the same thing for f.

    So it would be

    <br />
20 - \lim_{x \to -3}g(-3) = 13<br />


    <br />
\lim_{x \to -3}g(-3) = 7<br />
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by silencecloak View Post
    So it would be

    <br /> <br />
    " alt="20 - \lim_{x \to -3}g(-3) = 13
    " />

    <br /> <br />
    \lim_{x \to -3}g(-3) = 7<br /> <br />

    yes

    do you see why? do you understand what's going on? write out your full solution
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    yes

    do you see why? do you understand what's going on? write out your full solution
    <br />
\lim_{x \to -3} [4f(x)-g(x)] = 13<br />

    <br />
\lim_{x \to -3}4f(x) - \lim_{x \to -3}g(x) = 13<br />

    <br />
20 - \lim_{x \to -3}g(x) = 13<br />

    <br />
\lim_{x \to -3}g(x) = 7<br />
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by silencecloak View Post
    <br />
\lim_{x \to -3} [4f(x)-g(x)] = 13<br />

    <br />
\lim_{x \to -3}4f(x) - \lim_{x \to -3}g(x) = 13<br />

    <br />
20 - \lim_{x \to -3}g(x) = 13<br />

    <br />
\lim_{x \to -3}g(x) = 7<br />
    yes, and since g is continuous on the reals, \lim_{x \to -3}g(x) = g(-3), thus g(-3) = 7

    you must write that, otherwise, you have not answered the question. no one asked to for a limit, they asked you for g(-3)
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  11. #11
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    Quote Originally Posted by Jhevon View Post
    yes, and since g is continuous on the reals, \lim_{x \to -3}g(x) = g(-3), thus g(-3) = 7

    you must write that, otherwise, you have not answered the question. no one asked to for a limit, they asked you for g(-3)
    Oh ok, that one hit home for me

    Thanks for makin sure i understood it rather than just checking my answer
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by silencecloak View Post
    Oh ok, that one hit home for me

    Thanks for makin sure i understood it rather than just checking my answer
    i'd also put in an extra line to show that \lim_{x \to -3}4f(x) = 4f(-3). just so your professor knows you know what you're talking about
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