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The proofs of these properties are similar to those for the limits of functions of one variable. We can apply these laws to finding limits of various functions.

Finding the limit of a function of two variables

Find each of the following limits:

  1. lim ( x , y ) ( 2 , −1 ) ( x 2 2 x y + 3 y 2 4 x + 3 y 6 )
  2. lim ( x , y ) ( 2 , −1 ) 2 x + 3 y 4 x 3 y
  1. First use the sum and difference laws to separate the terms:
    lim ( x , y ) ( 2 , −1 ) ( x 2 2 x y + 3 y 2 4 x + 3 y 6 ) = ( lim ( x , y ) ( 2 , −1 ) x 2 ) ( lim ( x , y ) ( 2 , −1 ) 2 x y ) + ( lim ( x , y ) ( 2 , −1 ) 3 y 2 ) ( lim ( x , y ) ( 2 , −1 ) 4 x ) + ( lim ( x , y ) ( 2 , −1 ) 3 y ) ( lim ( x , y ) ( 2 , −1 ) 6 ) .

    Next, use the constant multiple law on the second, third, fourth, and fifth limits:
    = ( lim ( x , y ) ( 2 , −1 ) x 2 ) 2 ( lim ( x , y ) ( 2 , −1 ) x y ) + 3 ( lim ( x , y ) ( 2 , −1 ) y 2 ) 4 ( lim ( x , y ) ( 2 , −1 ) x ) + 3 ( lim ( x , y ) ( 2 , −1 ) y ) lim ( x , y ) ( 2 , −1 ) 6 .

    Now, use the power law on the first and third limits, and the product law on the second limit:
    = ( lim ( x , y ) ( 2 , −1 ) x ) 2 2 ( lim ( x , y ) ( 2 , −1 ) x ) ( lim ( x , y ) ( 2 , −1 ) y ) + 3 ( lim ( x , y ) ( 2 , −1 ) y ) 2 4 ( lim ( x , y ) ( 2 , −1 ) x ) + 3 ( lim ( x , y ) ( 2 , −1 ) y ) lim ( x , y ) ( 2 , −1 ) 6 .

    Last, use the identity laws on the first six limits and the constant law on the last limit:
    lim ( x , y ) ( 2 , −1 ) ( x 2 2 x y + 3 y 2 4 x + 3 y 6 ) = ( 2 ) 2 2 ( 2 ) ( −1 ) + 3 ( −1 ) 2 4 ( 2 ) + 3 ( −1 ) 6 = −6.
  2. Before applying the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, constant multiple law, and identity law,
    lim ( x , y ) ( 2 , −1 ) ( 4 x 3 y ) = lim ( x , y ) ( 2 , −1 ) 4 x lim ( x , y ) ( 2 , −1 ) 3 y = 4 ( lim ( x , y ) ( 2 , −1 ) x ) 3 ( lim ( x , y ) ( 2 , −1 ) y ) = 4 ( 2 ) 3 ( −1 ) = 11.

    Since the limit of the denominator is nonzero, the quotient law applies. We now calculate the limit of the numerator using the difference law, constant multiple law, and identity law:
    lim ( x , y ) ( 2 , −1 ) ( 2 x + 3 y ) = lim ( x , y ) ( 2 , −1 ) 2 x + lim ( x , y ) ( 2 , −1 ) 3 y = 2 ( lim ( x , y ) ( 2 , −1 ) x ) + 3 ( lim ( x , y ) ( 2 , −1 ) y ) = 2 ( 2 ) + 3 ( −1 ) = 1.

    Therefore, according to the quotient law we have
    lim ( x , y ) ( 2 , −1 ) 2 x + 3 y 4 x 3 y = lim ( x , y ) ( 2 , −1 ) ( 2 x + 3 y ) lim ( x , y ) ( 2 , −1 ) ( 4 x 3 y ) = 1 11 .
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Evaluate the following limit:

lim ( x , y ) ( 5 , −2 ) x 2 y y 2 + x 1 3 .

lim ( x , y ) ( 5 , −2 ) x 2 y y 2 + x 1 3 = 3 2

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Since we are taking the limit of a function of two variables, the point ( a , b ) is in 2 , and it is possible to approach this point from an infinite number of directions. Sometimes when calculating a limit, the answer varies depending on the path taken toward ( a , b ) . If this is the case, then the limit fails to exist. In other words, the limit must be unique, regardless of path taken.

Limits that fail to exist

Show that neither of the following limits exist:

  1. lim ( x , y ) ( 0 , 0 ) 2 x y 3 x 2 + y 2
  2. lim ( x , y ) ( 0 , 0 ) 4 x y 2 x 2 + 3 y 4
  1. The domain of the function f ( x , y ) = 2 x y 3 x 2 + y 2 consists of all points in the x y -plane except for the point ( 0 , 0 ) ( [link] ). To show that the limit does not exist as ( x , y ) approaches ( 0 , 0 ) , we note that it is impossible to satisfy the definition of a limit of a function of two variables because of the fact that the function takes different values along different lines passing through point ( 0 , 0 ) . First, consider the line y = 0 in the x y -plane. Substituting y = 0 into f ( x , y ) gives
    f ( x , 0 ) = 2 x ( 0 ) 3 x 2 + 0 2 = 0

    for any value of x . Therefore the value of f remains constant for any point on the x -axis, and as y approaches zero, the function remains fixed at zero.
    Next, consider the line y = x . Substituting y = x into f ( x , y ) gives
    f ( x , x ) = 2 x ( x ) 3 x 2 + x 2 = 2 x 2 4 x 2 = 1 2 .

    This is true for any point on the line y = x . If we let x approach zero while staying on this line, the value of the function remains fixed at 1 2 , regardless of how small x is.
    Choose a value for ε that is less than 1 / 2 —say, 1 / 4 . Then, no matter how small a δ disk we draw around ( 0 , 0 ) , the values of f ( x , y ) for points inside that δ disk will include both 0 and 1 2 . Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.
    In xyz space, the function f(x, y) = 2xy/(3x2 + y2) is shown, which is a slightly twisted plane, with values of 0 along the line y = 0 and values of ½ along the line y = x.
    Graph of the function f ( x , y ) = ( 2 x y ) / ( 3 x 2 + y 2 ) . Along the line y = 0 , the function is equal to zero; along the line y = x , the function is equal to 1 2 .

    In a similar fashion to a., we can approach the origin along any straight line passing through the origin. If we try the x -axis (i.e., y = 0 ) , then the function remains fixed at zero. The same is true for the y -axis. Suppose we approach the origin along a straight line of slope k . The equation of this line is y = k x . Then the limit becomes
    lim ( x , y ) ( 0 , 0 ) 4 x y 2 x 2 + 3 y 4 = lim ( x , y ) ( 0 , 0 ) 4 x ( k x ) 2 x 2 + 3 ( k x ) 4 = lim ( x , y ) ( 0 , 0 ) 4 k 2 x 3 x 2 + 3 k 4 x 4 = lim ( x , y ) ( 0 , 0 ) 4 k 2 x 1 + 3 k 4 x 2 = lim ( x , y ) ( 0 , 0 ) ( 4 k 2 x ) lim ( x , y ) ( 0 , 0 ) ( 1 + 3 k 4 x 2 ) = 0

    regardless of the value of k . It would seem that the limit is equal to zero. What if we chose a curve passing through the origin instead? For example, we can consider the parabola given by the equation x = y 2 . Substituting y 2 in place of x in f ( x , y ) gives
    lim ( x , y ) ( 0 , 0 ) 4 x y 2 x 2 + 3 y 4 = lim ( x , y ) ( 0 , 0 ) 4 ( y 2 ) y 2 ( y 2 ) 2 + 3 y 4 = lim ( x , y ) ( 0 , 0 ) 4 y 4 y 4 + 3 y 4 = lim ( x , y ) ( 0 , 0 ) 1 = 1.

    By the same logic in a., it is impossible to find a δ disk around the origin that satisfies the definition of the limit for any value of ε < 1 . Therefore, lim ( x , y ) ( 0 , 0 ) 4 x y 2 x 2 + 3 y 4 does not exist.
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Practice Key Terms 8

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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