ਸਿੱਖਿਆ ਦੇ ਉਦੇਸ਼
4.2.1 ਦੋ ਚਲਾਂ ਦੇ ਫਲਨ ਦੀ ਸੀਮਾ ਦੀ ਗਣਨਾ ਕਰੋ।
4.2.2 ਸਿੱਖੋ ਕਿ ਦੋ ਚਲਾਂ ਦਾ ਫਲਨ ਕਿਸੇ ਸੀਮਾ ਬਿੰਦੂ 'ਤੇ ਪਹੁੰਚਣ ਦੇ ਮਾਰਗ ਦੇ ਆਧਾਰ 'ਤੇ ਵੱਖ-ਵੱਖ ਮੁੱਲਾਂ ਤੱਕ ਕਿਵੇਂ ਪਹੁੰਚ ਸਕਦਾ ਹੈ।
4.2.3 ਦੋ ਚਲਾਂ ਦੇ ਫਲਨ ਦੀ ਨਿਰੰਤਰਤਾ ਲਈ ਸ਼ਰਤਾਂ ਦੱਸੋ।
4.2.4 ਕਿਸੇ ਬਿੰਦੂ 'ਤੇ ਦੋ ਚਲਾਂ ਦੇ ਫਲਨ ਦੀ ਨਿਰੰਤਰਤਾ ਦੀ ਪੁਸ਼ਟੀ ਕਰੋ।
4.2.5 ਤਿੰਨ ਜਾਂ ਵੱਧ ਚਲਾਂ ਦੇ ਫਲਨ ਦੀ ਸੀਮਾ ਦੀ ਗਣਨਾ ਕਰੋ ਅਤੇ ਕਿਸੇ ਬਿੰਦੂ 'ਤੇ ਫਲਨ ਦੀ ਨਿਰੰਤਰਤਾ ਦੀ ਪੁਸ਼ਟੀ ਕਰੋ।
ਅਸੀਂ ਹੁਣ ਇੱਕ ਤੋਂ ਵੱਧ ਚਲਾਂ ਦੇ ਫਲਨਾਂ ਦੀ ਜਾਂਚ ਕੀਤੀ ਹੈ ਅਤੇ ਦੇਖਿਆ ਹੈ ਕਿ ਉਨ੍ਹਾਂ ਨੂੰ ਕਿਵੇਂ ਗ੍ਰਾਫ ਕਰਨਾ ਹੈ। ਇਸ ਭਾਗ ਵਿੱਚ, ਅਸੀਂ ਦੇਖਦੇ ਹਾਂ ਕਿ ਇੱਕ ਤੋਂ ਵੱਧ ਚਲਾਂ ਦੇ ਫਲਨ ਦੀ ਸੀਮਾ ਕਿਵੇਂ ਲਈ ਜਾਂਦੀ ਹੈ, ਅਤੇ ਕਿਸੇ ਬਿੰਦੂ 'ਤੇ ਉਸਦੇ ਡੋਮੇਨ ਵਿੱਚ ਇੱਕ ਤੋਂ ਵੱਧ ਚਲਾਂ ਦੇ ਫਲਨ ਦਾ ਨਿਰੰਤਰ ਹੋਣ ਦਾ ਕੀ ਮਤਲਬ ਹੈ। ਇਹ ਪਤਾ ਲੱਗਦਾ ਹੈ ਕਿ ਇਹਨਾਂ ਸੰਕਲਪਾਂ ਦੇ ਅਜਿਹੇ ਪਹਿਲੂ ਹਨ ਜੋ ਇੱਕ ਚਲ ਦੇ ਫਲਨਾਂ ਨਾਲ ਨਹੀਂ ਵਾਪਰਦੇ।
ਦੋ ਚਲਾਂ ਦੇ ਫਲਨ ਦੀ ਸੀਮਾ
ਇੱਕ ਚਲ ਦੇ ਫਲਨ ਦੀ ਸੀਮਾ ਦੀ ਪਰਿਭਾਸ਼ਾ ਤੋਂ ਯਾਦ ਕਰੋ:
ਮੰਨ ਲਓ f(x) ਕਿਸੇ ਖੁੱਲ੍ਹੇ ਅੰਤਰਾਲ ਵਿੱਚ a ≠ x ਲਈ ਪਰਿਭਾਸ਼ਿਤ ਹੈ ਜਿਸ ਵਿੱਚ a ਸ਼ਾਮਲ ਹੈ। ਮੰਨ ਲਓ L ਕੋਈ ਅਸਲ ਸੰਖਿਆ ਹੈ। ਫਿਰ
ਜੇ ਹਰ ε > 0 ਲਈ, ਕੋਈ δ > 0 ਮੌਜੂਦ ਹੈ, ਜਿਸ ਨਾਲ ਜੇ 0 < |x - a| < δ ਹੋਵੇ, f ਦੇ ਡੋਮੇਨ ਵਿੱਚ ਸਾਰੇ x ਲਈ, ਤਾਂ
ਦੋ ਚਲਾਂ ਦੇ ਫਲਨ ਦੀ ਸੀਮਾ ਨੂੰ ਪਰਿਭਾਸ਼ਿਤ ਕਰਨ ਲਈ ਇਸ ਪਰਿਭਾਸ਼ਾ ਨੂੰ ਅਨੁਕੂਲ ਬਣਾਉਣ ਤੋਂ ਪਹਿਲਾਂ, ਸਾਨੂੰ ਪਹਿਲਾਂ ਇਹ ਦੇਖਣ ਦੀ ਲੋੜ ਹੈ ਕਿ ਇੱਕ ਚਲ ਵਿੱਚ ਖੁੱਲ੍ਹੇ ਅੰਤਰਾਲ ਦੇ ਵਿਚਾਰ ਨੂੰ ਦੋ ਚਲਾਂ ਵਿੱਚ ਖੁੱਲ੍ਹੇ ਅੰਤਰਾਲ ਤੱਕ ਕਿਵੇਂ ਵਧਾਇਆ ਜਾਵੇ।
ਪਰਿਭਾਸ਼ਾ
ਬਿੰਦੂ (a,b) ∈ ℝ² 'ਤੇ ਵਿਚਾਰ ਕਰੋ। ਬਿੰਦੂ (a,b) 'ਤੇ ਕੇਂਦਰਿਤ ਇੱਕ δ ਡਿਸਕ ਨੂੰ ਬਿੰਦੂ (a,b) 'ਤੇ ਕੇਂਦਰਿਤ δ ਰੇਡੀਅਸ ਵਾਲੀ ਇੱਕ ਖੁੱਲ੍ਹੀ ਡਿਸਕ ਵਜੋਂ ਪਰਿਭਾਸ਼ਿਤ ਕੀਤਾ ਗਿਆ ਹੈ—ਯਾਨੀ,
ਜਿਵੇਂ ਕਿ ਹੇਠਾਂ ਦਿੱਤੇ ਗ੍ਰਾਫ ਵਿੱਚ ਦਿਖਾਇਆ ਗਿਆ ਹੈ।
ਚਿੱਤਰ 4.14: ਬਿੰਦੂ (2,1) ਦੇ ਆਲੇ-ਦੁਆਲੇ ਇੱਕ δ ਡਿਸਕ।
δ ਡਿਸਕ ਦਾ ਵਿਚਾਰ ਦੋ ਚਲਾਂ ਦੇ ਫਲਨ ਦੀ ਸੀਮਾ ਦੀ ਪਰਿਭਾਸ਼ਾ ਵਿੱਚ ਪ੍ਰਗਟ ਹੁੰਦਾ ਹੈ। ਜੇ δ ਛੋਟਾ ਹੈ, ਤਾਂ δ ਡਿਸਕ ਵਿੱਚ ਸਾਰੇ ਬਿੰਦੂ (x,y) (a,b) ਦੇ ਨੇੜੇ ਹੁੰਦੇ ਹਨ। ਇਹ ਇੱਕ ਚਲ ਦੇ ਫਲਨ ਦੀ ਸੀਮਾ ਦੀ ਪਰਿਭਾਸ਼ਾ ਵਿੱਚ x ਦੇ a ਦੇ ਨੇੜੇ ਹੋਣ ਦੇ ਬਿਲਕੁਲ ਸਮਾਨ ਹੈ। ਇੱਕ ਅਯਾਮ ਵਿੱਚ, ਅਸੀਂ ਇਸ ਪਾਬੰਦੀ ਨੂੰ ਇਸ ਤਰ੍ਹਾਂ ਪ੍ਰਗਟ ਕਰਦੇ ਹਾਂ:
ਇੱਕ ਤੋਂ ਵੱਧ ਅਯਾਮਾਂ ਵਿੱਚ, ਅਸੀਂ ਇੱਕ δ ਡਿਸਕ ਦੀ ਵਰਤੋਂ ਕਰਦੇ ਹਾਂ।
ਪਰਿਭਾਸ਼ਾ
ਮੰਨ ਲਓ f ਦੋ ਚਲਾਂ, x ਅਤੇ y, ਦਾ ਇੱਕ ਫਲਨ ਹੈ। ਜਦੋਂ (x,y) (a,b) ਤੱਕ ਪਹੁੰਚਦਾ ਹੈ ਤਾਂ f(x,y) ਦੀ ਸੀਮਾ L ਹੈ, ਜਿਸਨੂੰ ਲਿਖਿਆ ਜਾਂਦਾ ਹੈ:
ਜੇ ਹਰ ε > 0 ਲਈ ਕੋਈ ਕਾਫ਼ੀ ਛੋਟਾ δ > 0 ਮੌਜੂਦ ਹੈ ਜਿਸ ਨਾਲ (a,b) ਦੇ ਆਲੇ-ਦੁਆਲੇ ਇੱਕ δ ਡਿਸਕ ਵਿੱਚ ਸਾਰੇ ਬਿੰਦੂਆਂ (x,y) ਲਈ, ਸੰਭਵ ਤੌਰ 'ਤੇ (a,b) ਨੂੰ ਛੱਡ ਕੇ, f(x,y) ਦਾ ਮੁੱਲ L ਤੋਂ ε ਤੋਂ ਵੱਧ ਦੂਰ ਨਹੀਂ ਹੈ (ਚਿੱਤਰ 4.15)। ਚਿੰਨ੍ਹਾਂ ਦੀ ਵਰਤੋਂ ਕਰਦੇ ਹੋਏ, ਅਸੀਂ ਹੇਠਾਂ ਲਿਖਦੇ ਹਾਂ: ਕਿਸੇ ਵੀ ε > 0 ਲਈ, ਕੋਈ ਸੰਖਿਆ δ > 0 ਮੌਜੂਦ ਹੈ ਜਿਸ ਨਾਲ
ਚਿੱਤਰ 4.15: ਦੋ ਚਲਾਂ ਨਾਲ ਸਬੰਧਤ ਫਲਨ ਦੀ ਸੀਮਾ ਲਈ f(x,y) ਦਾ L ਤੋਂ ε ਦੇ ਅੰਦਰ ਹੋਣਾ ਜ਼ਰੂਰੀ ਹੈ ਜਦੋਂ ਵੀ (x,y) (a,b) ਦੇ δ ਦੇ ਅੰਦਰ ਹੋਵੇ। ε ਦਾ ਮੁੱਲ ਜਿੰਨਾ ਛੋਟਾ ਹੋਵੇਗਾ, δ ਦਾ ਮੁੱਲ ਵੀ ਉਨਾ ਹੀ ਛੋਟਾ ਹੋਵੇਗਾ।
ਦੋ ਚਲਾਂ ਦੇ ਫਲਨ ਦੀ ਸੀਮਾ ਦੀ ਪਰਿਭਾਸ਼ਾ ਦੀ ਵਰਤੋਂ ਕਰਕੇ ਸੀਮਾ ਸਾਬਤ ਕਰਨਾ ਚੁਣੌਤੀਪੂਰਨ ਹੋ ਸਕਦਾ ਹੈ। ਇਸ ਦੀ ਬਜਾਏ, ਅਸੀਂ ਹੇਠਾਂ ਦਿੱਤੇ ਥਿਊਰਮ ਦੀ ਵਰਤੋਂ ਕਰਦੇ ਹਾਂ, ਜੋ ਸਾਨੂੰ ਸੀਮਾਵਾਂ ਲੱਭਣ ਲਈ ਸ਼ਾਰਟਕੱਟ ਪ੍ਰਦਾਨ ਕਰਦਾ ਹੈ। ਇਸ ਥਿਊਰਮ ਵਿੱਚ ਫਾਰਮੂਲੇ ਸੀਮਾ ਕਾਨੂੰਨਾਂ ਥਿਊਰਮ ਵਿੱਚ ਫਾਰਮੂਲਿਆਂ ਦਾ ਇੱਕ ਵਿਸਥਾਰ ਹਨ।
ਥਿਊਰਮ 4.1
Limit laws for functions of two variables
Let f(x,y)f(x,y) and g(x,y)g(x,y) be defined for all (x,y)≠(a,b)(x,y)≠(a,b) in a neighborhood around (a,b),(a,b), and assume the neighborhood is contained completely inside the domain of f.f. Assume that LL and MM are real numbers such that lim(x,y)→(a,b)f(x,y)=Llim(x,y)→(a,b)f(x,y)=L and lim(x,y)→(a,b)g(x,y)=M,lim(x,y)→(a,b)g(x,y)=M, and let cc be a constant. Then each of the following statements holds:
Constant Law:
Identity Laws:
Sum Law:
Difference Law:
Constant Multiple Law:
Product Law:
Quotient Law:
Power Law:
for any positive integer n.n.
Root Law:
for all LL if nn is odd and positive, and for L≥0L≥0 if nn is even and positive provided that f(x,y)≥0f(x,y)≥0 for all (x,y)≠(a,b)(x,y)≠(a,b) in neighborhood of (a,b)(a,b).
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.
Example 4.8
Finding the Limit of a Function of Two Variables
Find each of the following limits:
lim(x,y)→(2,−1)(x2−2xy+3y2−4x+3y−6)lim(x,y)→(2,−1)(x2−2xy+3y2−4x+3y−6)
lim(x,y)→(2,−1)2x+3y4x−3ylim(x,y)→(2,−1)2x+3y4x−3y
Solution
First use the sum and difference laws to separate the terms: lim(x,y)→(2,−1)(x2−2xy+3y2−4x+3y−6)=(lim(x,y)→(2,−1)x2)−(lim(x,y)→(2,−1)2xy)+(lim(x,y)→(2,−1)3y2)−(lim(x,y)→(2,−1)4x)+(lim(x,y)→(2,−1)3y)−(lim(x,y)→(2,−1)6).lim(x,y)→(2,−1)(x2−2xy+3y2−4x+3y−6)=(lim(x,y)→(2,−1)x2)−(lim(x,y)→(2,−1)2xy)+(lim(x,y)→(2,−1)3y2)−(lim(x,y)→(2,−1)4x)+(lim(x,y)→(2,−1)3y)−(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)x2)−2(lim(x,y)→(2,−1)xy)+3(lim(x,y)→(2,−1)y2)−4(lim(x,y)→(2,−1)x)+3(lim(x,y)→(2,−1)y)−lim(x,y)→(2,−1)6.=(lim(x,y)→(2,−1)x2)−2(lim(x,y)→(2,−1)xy)+3(lim(x,y)→(2,−1)y2)−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.=(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)(x2−2xy+3y2−4x+3y−6)=(2)2−2(2)(−1)+3(−1)2−4(2)+3(−1)−6=−6.lim(x,y)→(2,−1)(x2−2xy+3y2−4x+3y−6)=(2)2−2(2)(−1)+3(−1)2−4(2)+3(−1)−6=−6.
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)(4x−3y)=lim(x,y)→(2,−1)4x−lim(x,y)→(2,−1)3y=4(lim(x,y)→(2,−1)x)−3(lim(x,y)→(2,−1)y)=4(2)−3(−1)=11.lim(x,y)→(2,−1)(4x−3y)=lim(x,y)→(2,−1)4x−lim(x,y)→(2,−1)3y=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)(2x+3y)=lim(x,y)→(2,−1)2x+lim(x,y)→(2,−1)3y=2(lim(x,y)→(2,−1)x)+3(lim(x,y)→(2,−1)y)=2(2)+3(−1)=1.lim(x,y)→(2,−1)(2x+3y)=lim(x,y)→(2,−1)2x+lim(x,y)→(2,−1)3y=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)2x+3y4x−3y=lim(x,y)→(2,−1)(2x+3y)lim(x,y)→(2,−1)(4x−3y)=111.lim(x,y)→(2,−1)2x+3y4x−3y=lim(x,y)→(2,−1)(2x+3y)lim(x,y)→(2,−1)(4x−3y)=111.
Checkpoint 4.6
Evaluate the following limit:
Since we are taking the limit of a function of two variables, the point (a,b)(a,b) is in ℝ2,ℝ2, 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).(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.
Example 4.9
Limits That Fail to Exist
Show that neither of the following limits exist:
lim(x,y)→(0,0)2xy3x2+y2lim(x,y)→(0,0)2xy3x2+y2
lim(x,y)→(0,0)4xy2x2+3y4lim(x,y)→(0,0)4xy2x2+3y4
Solution
The domain of the function f(x,y)=2xy3x2+y2f(x,y)=2xy3x2+y2 consists of all points in the xy-planexy-plane except for the point (0,0)(0,0) (Figure 4.16). To show that the limit does not exist as (x,y)(x,y) approaches (0,0),(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).(0,0). First, consider the line y=0y=0 in the xy-plane.xy-plane. Substituting y=0y=0 into f(x,y)f(x,y) gives f(x,0)=2x(0)3x2+02=0f(x,0)=2x(0)3x2+02=0 for any value of x.x. Therefore the value of ff remains constant for any point on the x-axis,x-axis, and as yy approaches zero, the function remains fixed at zero. Next, consider the line y=x.y=x. Substituting y=xy=x into f(x,y)f(x,y) gives f(x,x)=2x(x)3x2+x2=2x24x2=12.f(x,x)=2x(x)3x2+x2=2x24x2=12. This is true for any point on the line y=x.y=x. If we let xx approach zero while staying on this line, the value of the function remains fixed at 12,12, regardless of how small xx is. Choose a value for εε that is less than 1/21/2—say, 1/4.1/4. Then, no matter how small a δδ disk we draw around (0,0),(0,0), the values of f(x,y)f(x,y) for points inside that δδ disk will include both 00 and 12.12. Therefore, the definition of limit at a point is never satisfied and the limit fails to exist. Figure 4.16 Graph of the function f(x,y)=(2xy)/(3x2+y2).f(x,y)=(2xy)/(3x2+y2). Along the line y=0,y=0, the function is equal to zero; along the line y=x,y=x, the function is equal to 12.12. In a similar fashion to a., we can approach the origin along any straight line passing through the origin. If we try the x-axisx-axis (i.e., y=0),y=0), then the function remains fixed at zero. The same is true for the y-axis.y-axis. Suppose we approach the origin along a straight line of slope k.k. The equation of this line is y=kx.y=kx. Then the limit becomes lim(x,y)→(0,0)4xy2x2+3y4=lim(x,y)→(0,0)4x(kx)2x2+3(kx)4=lim(x,y)→(0,0)4k2x3x2+3k4x4=lim(x,y)→(0,0)4k2x1+3k4x2=lim(x,y)→(0,0)(4k2x)lim(x,y)→(0,0)(1+3k4x2)=0lim(x,y)→(0,0)4xy2x2+3y4=lim(x,y)→(0,0)4x(kx)2x2+3(kx)4=lim(x,y)→(0,0)4k2x3x2+3k4x4=lim(x,y)→(0,0)4k2x1+3k4x2=lim(x,y)→(0,0)(4k2x)lim(x,y)→(0,0)(1+3k4x2)=0 regardless of the value of k.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=y2.x=y2. Substituting y2y2 in place of xx in f(x,y)f(x,y) gives lim(x,y)→(0,0)4xy2x2+3y4=lim(x,y)→(0,0)4(y2)y2(y2)2+3y4=lim(x,y)→(0,0)4y4y4+3y4=lim(x,y)→(0,0)1=1.lim(x,y)→(0,0)4xy2x2+3y4=lim(x,y)→(0,0)4(y2)y2(y2)2+3y4=lim(x,y)→(0,0)4y4y4+3y4=lim(x,y)→(0,0)1=1. By the same logic in a., it is impossible to find a δδ disk around the origin that satisfies the definition of the limit for any value of ε<1.ε<1. Therefore, lim(x,y)→(0,0)4xy2x2+3y4lim(x,y)→(0,0)4xy2x2+3y4 does not exist.
Checkpoint 4.7
Show that
does not exist.
Interior Points and Boundary Points
To study continuity and differentiability of a function of two or more variables, we first need to learn some new terminology.
Definition
Let S be a subset of ℝ2ℝ2 (Figure 4.17).
A point P0P0 is called an interior point of SS if there is a δδ disk centered around P0P0 contained completely in S.S.
A point P0P0 is called a boundary point of SS if every δδ disk centered around P0P0 contains points both inside and outside S.S.
Figure 4.17: In the set SS shown, (−1,1)(−1,1) is an interior point and (2,3)(2,3) is a boundary point.
Definition
Let S be a subset of ℝ2ℝ2 (Figure 4.17).
SS is called an open set if every point of SS is an interior point.
SS is called a closed set if it contains all its boundary points.
An example of an open set is a δδ disk. If we include the boundary of the disk, then it becomes a closed set. A set that contains some, but not all, of its boundary points is neither open nor closed. For example if we include half the boundary of a δδ disk but not the other half, then the set is neither open nor closed.
Definition
Let S be a subset of ℝ2ℝ2 (Figure 4.17).
An open set SS is a connected set if it cannot be represented as the union of two or more disjoint, nonempty open subsets.
A set SS is a region if it is open, connected, and nonempty.
The definition of a limit of a function of two variables requires the δδ disk to be contained inside the domain of the function. However, if we wish to find the limit of a function at a boundary point of the domain, the δdiskδdisk is not contained inside the domain. By definition, some of the points of the δdiskδdisk are inside the domain and some are outside. Therefore, we need only consider points that are inside both the δδ disk and the domain of the function. This leads to the definition of the limit of a function at a boundary point.
Definition
Let ff be a function of two variables, xx and y,y, and suppose (a,b)(a,b) is on the boundary of the domain of f.f. Then, the limit of f(x,y)f(x,y) as (x,y)(x,y) approaches (a,b)(a,b) is L,L, written
if for any ε>0,ε>0, there exists a number δ>0δ>0 such that for any point (x,y)(x,y) inside the domain of ff and within a suitably small distance positive δδ of (a,b),(a,b), the value of f(x,y)f(x,y) is no more than εε away from LL (Figure 4.15). Using symbols, we can write: For any ε>0,ε>0, there exists a number δ>0δ>0 such that
Example 4.10
Limit of a Function at a Boundary Point
Prove lim(x,y)→(4,3)25−x2−y2=0.lim(x,y)→(4,3)25−x2−y2=0.
Solution
The domain of the function f(x,y)=25−x2−y2f(x,y)=25−x2−y2 is {(x,y)∈ℝ2|x2+y2≤25},{(x,y)∈ℝ2|x2+y2≤25}, which is a circle of radius 55 centered at the origin, along with its interior as shown in the following graph.
Figure 4.18: Domain of the function f(x,y)=25−x2−y2.f(x,y)=25−x2−y2.
We can use the limit laws, which apply to limits at the boundary of domains as well as interior points:
See the following graph.
Figure 4.19: Graph of the function f(x,y)=25−x2−y2.f(x,y)=25−x2−y2.
Checkpoint 4.8
Evaluate the following limit:
Continuity of Functions of Two Variables
In Continuity, we defined the continuity of a function of one variable and saw how it relied on the limit of a function of one variable. In particular, three conditions are necessary for f(x)f(x) to be continuous at point x=a:x=a:
f(a)f(a) exists.
limx→af(x)limx→af(x) exists.
limx→af(x)=f(a).limx→af(x)=f(a).
These three conditions are necessary for continuity of a function of two variables as well.
Definition
A function f(x,y)f(x,y) is continuous at a point (a,b)(a,b) in its domain if the following conditions are satisfied:
f(a,b)f(a,b) exists.
lim(x,y)→(a,b)f(x,y)lim(x,y)→(a,b)f(x,y) exists.
lim(x,y)→(a,b)f(x,y)=f(a,b).lim(x,y)→(a,b)f(x,y)=f(a,b).
Example 4.11
Demonstrating Continuity for a Function of Two Variables
Show that the function f(x,y)=3x+2yx+y+1f(x,y)=3x+2yx+y+1 is continuous at point (5,−3).(5,−3).
Solution
There are three conditions to be satisfied, per the definition of continuity. In this example, a=5a=5 and b=−3.b=−3.
f(a,b)f(a,b) exists. This is true because the domain of the function ff consists of those ordered pairs for which the denominator is nonzero (i.e., x+y+1≠0).x+y+1≠0). Point (5,−3)(5,−3) satisfies this condition. Furthermore, f(a,b)=f(5,−3)=3(5)+2(−3)5+(−3)+1=15−62+1=3.f(a,b)=f(5,−3)=3(5)+2(−3)5+(−3)+1=15−62+1=3.
lim(x,y)→(a,b)f(x,y)lim(x,y)→(a,b)f(x,y) exists. This is also true: lim(x,y)→(a,b)f(x,y)=lim(x,y)→(5,−3)3x+2yx+y+1=lim(x,y)→(5,−3)(3x+2y)lim(x,y)→(5,−3)(x+y+1)=15−65−3+1=3.lim(x,y)→(a,b)f(x,y)=lim(x,y)→(5,−3)3x+2yx+y+1=lim(x,y)→(5,−3)(3x+2y)lim(x,y)→(5,−3)(x+y+1)=15−65−3+1=3.
lim(x,y)→(a,b)f(x,y)=f(a,b).lim(x,y)→(a,b)f(x,y)=f(a,b). This is true because we have just shown that both sides of this equation equal three.
Checkpoint 4.9
Show that the function f(x,y)=26−2x2−y2f(x,y)=26−2x2−y2 is continuous at point (2,−3).(2,−3).
Continuity of a function of any number of variables can also be defined in terms of delta and epsilon. A function of two variables is continuous at a point (x0,y0)(x0,y0) in its domain if for every ε>0ε>0 there exists a δ>0δ>0 such that, whenever (x−x0)2+(y−y0)2<δ(x−x0)2+(y−y0)2<δ it is true, |f(x,y)−f(a,b)|<ε.|f(x,y)−f(a,b)|<ε. This definition can be combined with the formal definition (that is, the epsilon–delta definition) of continuity of a function of one variable to prove the following theorems:
Theorem 4.2
The Sum of Continuous Functions Is Continuous
If f(x,y)f(x,y) is continuous at (x0,y0),(x0,y0), and g(x,y)g(x,y) is continuous at (x0,y0),(x0,y0), then f(x,y)+g(x,y)f(x,y)+g(x,y) is continuous at (x0,y0).(x0,y0).
Theorem 4.3
The Product of Continuous Functions Is Continuous
If g(x)g(x) is continuous at x0x0 and h(y)h(y) is continuous at y0,y0, then f(x,y)=g(x)h(y)f(x,y)=g(x)h(y) is continuous at (x0,y0).(x0,y0).
Theorem 4.4
The Composition of Continuous Functions Is Continuous
Let gg be a function of two variables from a domain D⊆ℝ2D⊆ℝ2 to a range R⊆ℝ.R⊆ℝ. Suppose gg is continuous at some point (x0,y0)∈D(x0,y0)∈D and define z0=g(x0,y0).z0=g(x0,y0). Let ff be a function that maps ℝℝ to ℝℝ such that z0z0 is in the domain of f.f. Last, assume ff is continuous at z0.z0. Then f∘gf∘g is continuous at (x0,y0)(x0,y0) as shown in the following figure.
Figure 4.20: The composition of two continuous functions is continuous.
Let’s now use the previous theorems to show continuity of functions in the following examples.
Example 4.12
More Examples of Continuity of a Function of Two Variables
Show that the functions f(x,y)=4x3y2f(x,y)=4x3y2 and g(x,y)=cos(4x3y2)g(x,y)=cos(4x3y2) are continuous everywhere.
Solution
The polynomials g(x)=4x3g(x)=4x3 and h(y)=y2h(y)=y2 are continuous at every real number, and therefore by the product of continuous functions theorem, f(x,y)=4x3y2f(x,y)=4x3y2 is continuous at every point (x,y)(x,y) in the xy-plane.xy-plane. Since f(x,y)=4x3y2f(x,y)=4x3y2 is continuous at every point (x,y)(x,y) in the xy-planexy-plane and g(x)=cosxg(x)=cosx is continuous at every real number x,x, the continuity of the composition of functions tells us that g(x,y)=cos(4x3y2)g(x,y)=cos(4x3y2) is continuous at every point (x,y)(x,y) in the xy-plane.xy-plane.
Checkpoint 4.10
Show that the functions f(x,y)=2x2y3+3f(x,y)=2x2y3+3 and g(x,y)=(2x2y3+3)4g(x,y)=(2x2y3+3)4 are continuous everywhere.
Functions of Three or More Variables
The limit of a function of three or more variables occurs readily in applications. For example, suppose we have a function f(x,y,z)f(x,y,z) that gives the temperature at a physical location (x,y,z)(x,y,z) in three dimensions. Or perhaps a function g(x,y,z,t)g(x,y,z,t) can indicate air pressure at a location (x,y,z)(x,y,z) at time t.t. How can we take a limit at a point in ℝ3?ℝ3? What does it mean to be continuous at a point in four dimensions?
The answers to these questions rely on extending the concept of a δδ disk into more than two dimensions. Then, the ideas of the limit of a function of three or more variables and the continuity of a function of three or more variables are very similar to the definitions given earlier for a function of two variables.
Definition
Let (x0,y0,z0)(x0,y0,z0) be a point in ℝ3.ℝ3. Then, a δδ ball in three dimensions consists of all points in ℝ3ℝ3 lying at a distance of less than δδ from (x0,y0,z0)(x0,y0,z0)—that is,
To define a δδ ball in higher dimensions, add additional terms under the radical to correspond to each additional dimension. For example, given a point P=(w0,x0,y0,z0)P=(w0,x0,y0,z0) in ℝ4,ℝ4, a δδ ball around PP can be described by
To show that a limit of a function of three variables exists at a point (x0,y0,z0),(x0,y0,z0), it suffices to show that for any point in a δδ ball centered at (x0,y0,z0),(x0,y0,z0), the value of the function at that point is arbitrarily close to a fixed value (the limit value). All the limit laws for functions of two variables hold for functions of more than two variables as well.
Example 4.13
Finding the Limit of a Function of Three Variables
Find lim(x,y,z)→(4,1,−3)x2y−3z2x+5y−z.lim(x,y,z)→(4,1,−3)x2y−3z2x+5y−z.
Solution
Before we can apply the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, the identity law, and the constant law,
Since this is nonzero, we next find the limit of the numerator. Using the product law, difference law, constant multiple law, and identity law,
Last, applying the quotient law:
Checkpoint 4.11
Find lim(x,y,z)→(4,−1,3)13−x2−2y2+z2.lim(x,y,z)→(4,−1,3)13−x2−2y2+z2.
For the following exercises, find the limit of the function.
lim(x,y)→(1,2)xlim(x,y)→(1,2)x
lim(x,y)→(1,2)5x2yx2+y2lim(x,y)→(1,2)5x2yx2+y2
Show that the limit lim(x,y)→(0,0)5x2yx2+y2lim(x,y)→(0,0)5x2yx2+y2 exists and is the same along the paths: y-axisy-axis and x-axis,x-axis, and along y=x.y=x.
For the following exercises, evaluate the limits at the indicated values of xandy.xandy. If the limit does not exist, state this and explain why the limit does not exist.
lim(x,y)→(0,0)4x2+10y2+44x2−10y2+6lim(x,y)→(0,0)4x2+10y2+44x2−10y2+6
lim(x,y)→(11,13)1xylim(x,y)→(11,13)1xy
lim(x,y)→(0,1)y2sinxxlim(x,y)→(0,1)y2sinxx
lim(x,y)→(0,0)sin(x8+y7x−y+10)lim(x,y)→(0,0)sin(x8+y7x−y+10)
lim(x,y)→(π/4,1)ytanxy+1lim(x,y)→(π/4,1)ytanxy+1
lim(x,y)→(0,π/4)secx+23x−tanylim(x,y)→(0,π/4)secx+23x−tany
lim(x,y)→(2,5)(1x−5y)lim(x,y)→(2,5)(1x−5y)
lim(x,y)→(4,4)xlnylim(x,y)→(4,4)xlny
lim(x,y)→(4,4)e−x2−y2lim(x,y)→(4,4)e−x2−y2
lim(x,y)→(0,0)9−x2−y2lim(x,y)→(0,0)9−x2−y2
lim(x,y)→(1,2)(x2y3−x3y2+3x+2y)lim(x,y)→(1,2)(x2y3−x3y2+3x+2y)
lim(x,y)→(π,π)xsin(x+y4)lim(x,y)→(π,π)xsin(x+y4)
lim(x,y)→(0,0)xy+1x2+y2+1lim(x,y)→(0,0)xy+1x2+y2+1
lim(x,y)→(0,0)x2+y2x2+y2+1−1lim(x,y)→(0,0)x2+y2x2+y2+1−1
lim(x,y)→(0,0)ln(x2+y2)lim(x,y)→(0,0)ln(x2+y2)
For the following exercises, complete the statement.
A point (x0,y0)(x0,y0) in a plane region RR is an interior point of RR if _________________.
A point (x0,y0)(x0,y0) in a plane region RR is called a boundary point of RR if ___________.
For the following exercises, use algebraic techniques to evaluate the limit.
lim(x,y)→(2,1)x−y−1x−y−1lim(x,y)→(2,1)x−y−1x−y−1
lim(x,y)→(0,0)x4−4y4x2+2y2lim(x,y)→(0,0)x4−4y4x2+2y2
lim(x,y)→(0,0)x3−y3x−ylim(x,y)→(0,0)x3−y3x−y
lim(x,y)→(0,0)x2−xyx−ylim(x,y)→(0,0)x2−xyx−y
For the following exercises, evaluate the limits of the functions of three variables.
lim(x,y,z)→(1,2,3)xz2−y2zxyz−1lim(x,y,z)→(1,2,3)xz2−y2zxyz−1
lim(x,y,z)→(0,0,0)x2−y2−z2x2+y2−z2lim(x,y,z)→(0,0,0)x2−y2−z2x2+y2−z2
For the following exercises, evaluate the limit of the function by determining the value the function approaches along the indicated paths. If the limit does not exist, explain why not.
lim(x,y)→(0,0)xy+y3x2+y2lim(x,y)→(0,0)xy+y3x2+y2
Along the x-axisx-axis (y=0)(y=0)
Along the y-axisy-axis (x=0)(x=0)
Along the path y=2xy=2x
Evaluate lim(x,y)→(0,0)xy+y3x2+y2lim(x,y)→(0,0)xy+y3x2+y2 using the results of previous problem.
lim(x,y)→(0,0)x2yx4+y2lim(x,y)→(0,0)x2yx4+y2
Along the x-axis (y=0)(y=0)
Along the y-axis (x=0)(x=0)
Along the path y=x2y=x2
Evaluate lim(x,y)→(0,0)x2yx4+y2lim(x,y)→(0,0)x2yx4+y2 using the results of previous problem.
Discuss the continuity of the following functions. Find the largest region in the xy-planexy-plane in which the following functions are continuous.
f(x,y)=sin(xy)f(x,y)=sin(xy)
f(x,y)=ln(x+y)f(x,y)=ln(x+y)
f(x,y)=e3xyf(x,y)=e3xy
f(x,y)=1xyf(x,y)=1xy
For the following two exercises, determine the region in which the function is continuous. Explain your answer.
f(x,y)=x2yx2+y2f(x,y)=x2yx2+y2
f(x,y)=sin(x2+y2)x2+y2f(x,y)=sin(x2+y2)x2+y2
Determine whether fx,yfx,y is continuous at 0,00,0. f(x,y)={x2yx2+y2if(x,y)≠(0,0)0if(x,y)=(0,0)}f(x,y)={x2yx2+y2if(x,y)≠(0,0)0if(x,y)=(0,0)}
Determine whether g(x,y)=x2−y2x2+y2g(x,y)=x2−y2x2+y2 is continuous at (0,0).(0,0).
Create a plot using graphing software to determine where the limit does not exist. Find where in the coordinate plane f(x,y)=1x2−yf(x,y)=1x2−y is continuous.
Determine the region of the xy-planexy-plane in which the function g(x,y)=arctan(xy2x+y)g(x,y)=arctan(xy2x+y) is continuous. Use technology to support your conclusion.
Determine the region of the xy-planexy-plane in which f(x,y)=ln(x2+y2−1)f(x,y)=ln(x2+y2−1) is continuous. Use technology to support your conclusion. (Hint: Choose the range of values for xandyxandy carefully!)
At what points in space is g(x,y,z)=x2+y2−2z2g(x,y,z)=x2+y2−2z2 continuous?
At what points in space is g(x,y,z)=1x2+z2−1g(x,y,z)=1x2+z2−1 continuous?
Show that lim(x,y)→(0,0)1x2+y2lim(x,y)→(0,0)1x2+y2 does not exist at (0,0)(0,0) by plotting the graph of the function.
[T] Evaluate lim(x,y)→(0,0)−xy2x2+y4lim(x,y)→(0,0)−xy2x2+y4 by plotting the function using a CAS. Determine analytically the limit along the path x=y2.x=y2.
[T]
Use a CAS to draw a contour map of z=9−x2−y2.z=9−x2−y2.
What is the name of the geometric shape of the level curves?
Give the general equation of the level curves.
What is the maximum value of z?z?
What is the domain of the function?
What is the range of the function?
True or False: If we evaluate lim(x,y)→(0,0)f(x)lim(x,y)→(0,0)f(x) along several paths and each time the limit is 1,1, we can conclude that lim(x,y)→(0,0)f(x)=1.lim(x,y)→(0,0)f(x)=1.
Use polar coordinates to find lim(x,y)→(0,0)sinx2+y2x2+y2.lim(x,y)→(0,0)sinx2+y2x2+y2. You can also find the limit using L’Hôpital’s rule.
Use polar coordinates to find lim(x,y)→(0,0)cos(x2+y2).lim(x,y)→(0,0)cos(x2+y2).
Discuss the continuity of f(g(x,y))f(g(x,y)) where f(t)=1/tf(t)=1/t and g(x,y)=2x−5y.g(x,y)=2x−5y.
Given f(x,y)=x2−4y,f(x,y)=x2−4y, find limh→0f(x+h,y)−f(x,y)h.limh→0f(x+h,y)−f(x,y)h.
Given f(x,y)=x2−4y,f(x,y)=x2−4y, find limh→0f(1+h,y)−f(1,y)h.limh→0f(1+h,y)−f(1,y)h.