Results 1 to 5 of 5

Math Help - Solving limits by change of variable

  1. #1
    Junior Member
    Joined
    Jul 2007
    Posts
    45

    Solving limits by change of variable

    I learned today that one of the methods for solving a limit is by change of variable. However, I'm confused as to how it works. I have the following question as an example:

    Evaluate the limit by change of variable.

    lim x->8

    3√x - 2
    ______
    x-8

    I started by saying let 3√x = y. So the numerator becomes y-2. However, I'm uncertain how to change the denominator when solving by change of variable , and how to determine what the new lim y-> is (are there general rules for these?). Thank you so much for any help, as I have a test on this on Friday.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by starswept View Post
    I learned today that one of the methods for solving a limit is by change of variable. However, I'm confused as to how it works. I have the following question as an example:

    Evaluate the limit by change of variable.

    lim x->8

    3√x - 2
    ______
    x-8

    I started by saying let 3√x = y. So the numerator becomes y-2. However, I'm uncertain how to change the denominator when solving by change of variable , and how to determine what the new lim y-> is (are there general rules for these?). Thank you so much for any help, as I have a test on this on Friday.
    just to clarify, by 3√x do you mean 3 times the square root of x or the cube root of x?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2007
    Posts
    45
    Quote Originally Posted by Jhevon View Post
    just to clarify, by 3√x do you mean 3 times the square root of x or the cube root of x?
    Sorry about that, I mean the cube root of x
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by starswept View Post
    Sorry about that, I mean the cube root of x
    i am not familiar with this change of variable method you refer to. i'd do this problem by realizing the denominator is the difference of two cubes, of which one of the factors will cancel with a term in the top
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    Let \sqrt[3]{x}=y\Rightarrow x=y^3.
    If x\to 8 then y\to 2.
    So \displaystyle \lim_{x\to 8}\frac{\sqrt[3]{x}-2}{x-8}=\lim_{y\to 2}\frac{y-2}{y^3-8}=\lim_{y\to 2}\frac{y-2}{(y-2)(y^2+2y+4)}=
    \displaystyle =\lim_{y\to 2}\frac{1}{y^2+2y+4}=\frac{1}{12}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 3rd 2010, 08:08 PM
  2. Replies: 4
    Last Post: June 25th 2010, 07:37 AM
  3. Replies: 6
    Last Post: March 21st 2009, 09:51 AM
  4. Replies: 1
    Last Post: March 21st 2009, 08:43 AM
  5. Another limits by change of variable problem
    Posted in the Calculus Forum
    Replies: 6
    Last Post: July 12th 2007, 02:08 PM

Search Tags


/mathhelpforum @mathhelpforum