Results 1 to 7 of 7

Math Help - Difference between left and right derivatives?

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    3

    Difference between left and right derivatives?

    Hello,

    Would someone be able to provide me with an example to clarify the definitions of \mathop {\lim }\limits_{x \to {x_0} + } f\left( x \right) and \mathop {\lim }\limits_{x \to {x_0} - } f\left( x \right)? I believe that I understand the calculus theorems, but not this (seemingly basic) idea of left and right limits. Both that, and I can never remember which way round their definitions are!

    Sorry, I should have posted this in the analysis section. If someone could move it there, that would be great.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    I'm not sure what you're looking for...?

    The limit, as x approaches some value "a", of f(x) is nothing more than what the value of f(a) ought reasonably to be, assuming the limit exists.

    In some cases, the limit does not exist, such as for f(x) = 1/x when x approaches zero.

    In other cases, the limit exists, but not the functional value, such as for f(x) = [(x + 1)(x - 2)]/(x - 2) when x approaches 2. Other than for x = 2, this function is the same as g(x) = x + 1, so the limit at x = 2 is g = 3. But f(x) is not actually defined for x = 2. The limit exists, but the function doesn't actually take on that value.

    In still other cases, the limit exist, and the functional value exists, such as for f(x) = x.

    And then you have the cases where the limit from one side exists, but the other limit does not, or is not the same value. A piecewise function is a good example of this:

    f(x)\, =\, \left\{\begin{array}{rr}-1&\mbox{ for }\, x\, \leq\, 0\\1&\mbox{ for }\, x\, >\, 0\end{array}\right.

    Clearly, each "half" has a limit as x approaches zero: from the left, the function "ought" to take on the value -1 (and it does); from the right, the function "ought" to take on the value 1 (but it doesn't, because f(0) = -1). Each one-sided limit exists, but "the" limit does not, because the two one-sided limits don't agree.

    Does that help at all...?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2009
    Posts
    3
    Quote Originally Posted by stapel View Post
    I'm not sure what you're looking for...?

    The limit, as x approaches some value "a", of f(x) is nothing more than what the value of f(a) ought reasonably to be, assuming the limit exists.

    In some cases, the limit does not exist, such as for f(x) = 1/x when x approaches zero.

    In other cases, the limit exists, but not the functional value, such as for f(x) = [(x + 1)(x - 2)]/(x - 2) when x approaches 2. Other than for x = 2, this function is the same as g(x) = x + 1, so the limit at x = 2 is g = 3. But f(x) is not actually defined for x = 2. The limit exists, but the function doesn't actually take on that value.

    In still other cases, the limit exist, and the functional value exists, such as for f(x) = x.

    And then you have the cases where the limit from one side exists, but the other limit does not, or is not the same value. A piecewise function is a good example of this:

    f(x)\, =\, \left\{\begin{array}{rr}-1&\mbox{ for }\, x\, \leq\, 0\\1&\mbox{ for }\, x\, >\, 0\end{array}\right.

    Clearly, each "half" has a limit as x approaches zero: from the left, the function "ought" to take on the value -1 (and it does); from the right, the function "ought" to take on the value 1 (but it doesn't, because f(0) = -1). Each one-sided limit exists, but "the" limit does not, because the two one-sided limits don't agree.

    Does that help at all...?
    Yes, this helps a lot. Just one last thing: in the example you wrote above, which value is the left derivative, and which is the right derivative? Does the left derivative mean 'approaches from the left toward 0' or 'moving to the left towards 0'?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    Quote Originally Posted by teaspoon View Post
    Does the left derivative mean 'approaches from the left toward 0'...?
    Yes.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2009
    Posts
    4
    Quote Originally Posted by teaspoon View Post
    Yes, this helps a lot. Just one last thing: in the example you wrote above, which value is the left derivative, and which is the right derivative? Does the left derivative mean 'approaches from the left toward 0' or 'moving to the left towards 0'?
    The proper way of saying it would be "the limit as x approaches zero from the left" The derivative is a special kind of limit; and if the limit must exist at that point.

    One-sided limits are used to determine a few things; and they allow you to find a limit value even if the limit does not necessarily exist. Such a case would be testing limits at negative and positive infinity to test for horizontal asymptotes.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2009
    Posts
    3
    Quote Originally Posted by Raldikuk View Post
    The proper way of saying it would be "the limit as x approaches zero from the left" The derivative is a special kind of limit; and if the limit must exist at that point.

    One-sided limits are used to determine a few things; and they allow you to find a limit value even if the limit does not necessarily exist. Such a case would be testing limits at negative and positive infinity to test for horizontal asymptotes.
    So, in the example above, \mathop {\lim }\limits_{x \to 0 - } f\left( x \right) =  - 1 and \mathop {\lim }\limits_{x \to 0 + } f\left( x \right) = 1?

    And, since \mathop {\lim }\limits_{x \to 0 + } f\left( x \right) \ne \mathop {\lim }\limits_{x \to 0 - } f\left( x \right), then \mathop {\lim }\limits_{x \to 0} f\left( x \right) does not exist? (I'm fairly sure the second part about the inequality is correct, just clarifying.)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Apr 2009
    Posts
    4
    Quote Originally Posted by teaspoon View Post
    So, in the example above, \mathop {\lim }\limits_{x \to 0 - } f\left( x \right) =  - 1 and \mathop {\lim }\limits_{x \to 0 + } f\left( x \right) = 1?

    And, since \mathop {\lim }\limits_{x \to 0 + } f\left( x \right) \ne \mathop {\lim }\limits_{x \to 0 - } f\left( x \right), then \mathop {\lim }\limits_{x \to 0} f\left( x \right) does not exist? (I'm fairly sure the second part about the inequality is correct, just clarifying.)
    Exactly; the limit itself does not exist, since the approach doesn't lead to one conclusive value; however it is still useful to be able to analyze the approach from the left or from the right.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: March 31st 2011, 06:23 AM
  2. Left Inverse
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: February 14th 2011, 02:24 PM
  3. How many left?
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: August 24th 2010, 07:45 PM
  4. Left continuity
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: June 3rd 2009, 02:41 AM
  5. Left/right eigenvectors
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 18th 2009, 05:22 AM

Search Tags


/mathhelpforum @mathhelpforum