Limits and Continuity

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Find each of the following limits:

   1. \(\underset{(x,y)\to (2,-1)}{\text{lim}}({x}^{2}-2xy+3{y}^{2}-4x+3y-6)\)
   2. \(\underset{(x,y)\to (2,-1)}{\text{lim}}\frac{2x+3y}{4x-3y}\)
   जवाफ प्रकट गर्नुहोस्

   1. First use the sum and difference laws to separate the terms:
      \[\begin{array}{l} \\ \underset{(x,y)\to (2,-1)}{\text{lim}}({x}^{2}-2xy+3{y}^{2}-4x+3y-6) \\ =(\underset{(x,y)\to (2,-1)}{\text{lim}}{x}^{2})-(\underset{(x,y)\to (2,-1)}{\text{lim}}2xy)+(\underset{(x,y)\to (2,-1)}{\text{lim}}3{y}^{2})-(\underset{(x,y)\to (2,-1)}{\text{lim}}4x) \\ +(\underset{(x,y)\to (2,-1)}{\text{lim}}3y)-(\underset{(x,y)\to (2,-1)}{\text{lim}}6).\end{array}\]
      Next, use the constant multiple law on the second, third, fourth, and fifth limits:
      \[\begin{array}{l} \\ \\ =(\underset{(x,y)\to (2,-1)}{\text{lim}}{x}^{2})-2(\underset{(x,y)\to (2,-1)}{\text{lim}}xy)+3(\underset{(x,y)\to (2,-1)}{\text{lim}}{y}^{2})-4(\underset{(x,y)\to (2,-1)}{\text{lim}}x) \\ +3(\underset{(x,y)\to (2,-1)}{\text{lim}}y)-\underset{(x,y)\to (2,-1)}{\text{lim}}6.\end{array}\]
      Now, use the power law on the first and third limits, and the product law on the second limit:
      \[\begin{array}{l} \\ \\ \\ ={(\underset{(x,y)\to (2,-1)}{\text{lim}}x)}^{2}-2(\underset{(x,y)\to (2,-1)}{\text{lim}}x)(\underset{(x,y)\to (2,-1)}{\text{lim}}y)+3{(\underset{(x,y)\to (2,-1)}{\text{lim}}y)}^{2} \\ -4(\underset{(x,y)\to (2,-1)}{\text{lim}}x)+3(\underset{(x,y)\to (2,-1)}{\text{lim}}y)-\underset{(x,y)\to (2,-1)}{\text{lim}}6.\end{array}\]
      Last, use the identity laws on the first six limits and the constant law on the last limit:
      \(\begin{array}{ll}\underset{(x,y)\to (2,-1)}{\text{lim}}({x}^{2}-2xy+3{y}^{2}-4x+3y-6) & ={(2)}^{2}-2(2)(-1)+3{(-1)}^{2}-4(2)+3(-1)-6 \\ & =-6.\end{array}\)
   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,
      \[\begin{array}{ll}\underset{(x,y)\to (2,-1)}{\text{lim}}(4x-3y) & =\underset{(x,y)\to (2,-1)}{\text{lim}}4x-\underset{(x,y)\to (2,-1)}{\text{lim}}3y \\ & =4(\underset{(x,y)\to (2,-1)}{\text{lim}}x)-3(\underset{(x,y)\to (2,-1)}{\text{lim}}y) \\ & =4(2)-3(-1)=11.\end{array}\]
      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:
      \[\begin{array}{ll}\underset{(x,y)\to (2,-1)}{\text{lim}}(2x+3y) & =\underset{(x,y)\to (2,-1)}{\text{lim}}2x+\underset{(x,y)\to (2,-1)}{\text{lim}}3y \\ & =2(\underset{(x,y)\to (2,-1)}{\text{lim}}x)+3(\underset{(x,y)\to (2,-1)}{\text{lim}}y) \\ & =2(2)+3(-1) \\ & =1.\end{array}\]
      Therefore, according to the quotient law we have
      \[\underset{(x,y)\to (2,-1)}{\text{lim}}\frac{2x+3y}{4x-3y}=\frac{\underset{(x,y)\to (2,-1)}{\text{lim}}(2x+3y)}{\underset{(x,y)\to (2,-1)}{\text{lim}}(4x-3y)}=\frac{1}{11}.\]

2. Evaluate the following limit:

   \[\underset{(x,y)\to (5,-2)}{\text{lim}}\sqrt[3]{\frac{{x}^{2}-y}{{y}^{2}+x-1}}.\]
   जवाफ प्रकट गर्नुहोस्

   \(\underset{(x,y)\to (5,-2)}{\text{lim}}\sqrt[3]{\frac{{x}^{2}-y}{{y}^{2}+x-1}}=\frac{3}{2}\)

3. Show that neither of the following limits exist:

   1. \(\underset{(x,y)\to (0,0)}{\text{lim}}\frac{2xy}{3{x}^{2}+{y}^{2}}\)
   2. \(\underset{(x,y)\to (0,0)}{\text{lim}}\frac{4x{y}^{2}}{{x}^{2}+3{y}^{4}}\)
   जवाफ प्रकट गर्नुहोस्

   1. The domain of the function \(f(x,y)=\frac{2xy}{3{x}^{2}+{y}^{2}}\) consists of all points in the \(xy\text{-plane}\) except for the point \((0,0)\) (). 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 \(xy\text{-plane.}\) Substituting \(y=0\) into \(f(x,y)\) gives
      \[f(x,0)=\frac{2x(0)}{3{x}^{2}+{0}^{2}}=0\]
      for any value of \(x.\) Therefore the value of \(f\) remains constant for any point on the \(x\text{-axis,}\) 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
      \[f(x,x)=\frac{2x(x)}{3{x}^{2}+{x}^{2}}=\frac{2{x}^{2}}{4{x}^{2}}=\frac{1}{2}.\]
      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 \(\frac{1}{2},\) regardless of how small \(x\) is.
      Choose a value for \(\epsilon\) that is less than \(1\text{/}2\)—say, \(1\text{/}4.\) Then, no matter how small a \(\delta\) disk we draw around \((0,0),\) the values of \(f(x,y)\) for points inside that \(\delta\) disk will include both \(0\) and \(\frac{1}{2}.\) Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.
      
      In a similar fashion to a., we can approach the origin along any straight line passing through the origin. If we try the \(x\text{-axis}\) (i.e., \(y=0),\) then the function remains fixed at zero. The same is true for the \(y\text{-axis.}\) Suppose we approach the origin along a straight line of slope \(k.\) The equation of this line is \(y=kx.\) Then the limit becomes
      \[\begin{array}{ll}\underset{(x,y)\to (0,0)}{\text{lim}}\frac{4x{y}^{2}}{{x}^{2}+3{y}^{4}} & =\underset{(x,y)\to (0,0)}{\text{lim}}\frac{4x{(kx)}^{2}}{{x}^{2}+3{(kx)}^{4}} \\ & =\underset{(x,y)\to (0,0)}{\text{lim}}\frac{4{k}^{2}{x}^{3}}{{x}^{2}+3{k}^{4}{x}^{4}} \\ & =\underset{(x,y)\to (0,0)}{\text{lim}}\frac{4{k}^{2}x}{1+3{k}^{4}{x}^{2}} \\ & =\frac{\underset{(x,y)\to (0,0)}{\text{lim}}(4{k}^{2}x)}{\underset{(x,y)\to (0,0)}{\text{lim}}(1+3{k}^{4}{x}^{2})} \\ & =0\end{array}\]
      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
      \[\begin{array}{ll}\underset{(x,y)\to (0,0)}{\text{lim}}\frac{4x{y}^{2}}{{x}^{2}+3{y}^{4}} & =\underset{(x,y)\to (0,0)}{\text{lim}}\frac{4({y}^{2}){y}^{2}}{{({y}^{2})}^{2}+3{y}^{4}} \\ & =\underset{(x,y)\to (0,0)}{\text{lim}}\frac{4{y}^{4}}{{y}^{4}+3{y}^{4}} \\ & =\underset{(x,y)\to (0,0)}{\text{lim}}1 \\ & =1.\end{array}\]
      By the same logic in a., it is impossible to find a \(\delta\) disk around the origin that satisfies the definition of the limit for any value of \(\epsilon <1.\) Therefore, \(\underset{(x,y)\to (0,0)}{\text{lim}}\frac{4x{y}^{2}}{{x}^{2}+3{y}^{4}}\) does not exist.

4. Show that

   \[\underset{(x,y)\to (2,1)}{\text{lim}}\frac{(x-2)(y-1)}{{(x-2)}^{2}+{(y-1)}^{2}}\]

   does not exist.

   जवाफ प्रकट गर्नुहोस्

   If \(y=k(x-2)+1,\) then \(\underset{(x,y)\to (2,1)}{\text{lim}}\frac{(x-2)(y-1)}{{(x-2)}^{2}+{(y-1)}^{2}}=\frac{k}{1+{k}^{2}}.\) Since the answer depends on \(k,\) the limit fails to exist.

5. Prove \(\underset{(x,y)\to (4,3)}{\text{lim}}\sqrt{25-{x}^{2}-{y}^{2}}=0.\)

   जवाफ प्रकट गर्नुहोस्

   The domain of the function \(f(x,y)=\sqrt{25-{x}^{2}-{y}^{2}}\) is \(\{(x,y)\in {ℝ}^{2}|{x}^{2}+{y}^{2}\le 25\},\) which is a circle of radius \(5\) centered at the origin, along with its interior as shown in the following graph.

   We can use the limit laws, which apply to limits at the boundary of domains as well as interior points:

   \[\begin{array}{ll}\underset{(x,y)\to (4,3)}{\text{lim}}\sqrt{25-{x}^{2}-{y}^{2}} & =\sqrt{\underset{(x,y)\to (4,3)}{\text{lim}}(25-{x}^{2}-{y}^{2})} \\ & =\sqrt{\underset{(x,y)\to (4,3)}{\text{lim}}25-\underset{(x,y)\to (4,3)}{\text{lim}}{x}^{2}-\underset{(x,y)\to (4,3)}{\text{lim}}{y}^{2}} \\ & =\sqrt{25-{4}^{2}-{3}^{2}} \\ & =0.\end{array}\]

   See the following graph.

6. Evaluate the following limit:

   \[\underset{(x,y)\to (5,-2)}{\text{lim}}\sqrt{29-{x}^{2}-{y}^{2}}.\]
   जवाफ प्रकट गर्नुहोस्

   \(\underset{(x,y)\to (5,-2)}{\text{lim}}\sqrt{29-{x}^{2}-{y}^{2}}=0\)

7. Show that the function \(f(x,y)=\frac{3x+2y}{x+y+1}\) is continuous at point \((5,-3).\)

   जवाफ प्रकट गर्नुहोस्

   There are three conditions to be satisfied, per the definition of continuity. In this example, \(a=5\) and \(b=-3.\)

   1. \(f(a,b)\) exists. This is true because the domain of the function \(f\) consists of those ordered pairs for which the denominator is nonzero (i.e., \(x+y+1\ne 0).\) Point \((5,-3)\) satisfies this condition. Furthermore,
      \[f(a,b)=f(5,-3)=\frac{3(5)+2(-3)}{5+(-3)+1}=\frac{15-6}{2+1}=3.\]
   2. \(\underset{(x,y)\to (a,b)}{\text{lim}}f(x,y)\) exists. This is also true:
      \[\begin{array}{ll}\underset{(x,y)\to (a,b)}{\text{lim}}f(x,y) & =\underset{(x,y)\to (5,-3)}{\text{lim}}\frac{3x+2y}{x+y+1} \\ & =\frac{\underset{(x,y)\to (5,-3)}{\text{lim}}(3x+2y)}{\underset{(x,y)\to (5,-3)}{\text{lim}}(x+y+1)} \\ & =\frac{15-6}{5-3+1} \\ & =3.\end{array}\]
   3. \(\underset{(x,y)\to (a,b)}{\text{lim}}f(x,y)=f(a,b).\) This is true because we have just shown that both sides of this equation equal three.

8. Show that the function \(f(x,y)=\sqrt{26-2{x}^{2}-{y}^{2}}\) is continuous at point \((2,-3).\)

   जवाफ प्रकट गर्नुहोस्

   1. The domain of \(f\) contains the ordered pair \((2,-3)\) because \(f(a,b)=f(2,-3)=\sqrt{16-2{(2)}^{2}-{(-3)}^{2}}=3\)
   2. \(\underset{(x,y)\to (a,b)}{\text{lim}}f(x,y)=3\)
   3. \(\underset{(x,y)\to (a,b)}{\text{lim}}f(x,y)=f(a,b)=3\)

9. Show that the functions \(f(x,y)=4{x}^{3}{y}^{2}\) and \(g(x,y)=\text{cos}(4{x}^{3}{y}^{2})\) are continuous everywhere.

   जवाफ प्रकट गर्नुहोस्

   The polynomials \(g(x)=4{x}^{3}\) and \(h(y)={y}^{2}\) are continuous at every real number, and therefore by the product of continuous functions theorem, \(f(x,y)=4{x}^{3}{y}^{2}\) is continuous at every point \((x,y)\) in the \(xy\text{-plane.}\) Since \(f(x,y)=4{x}^{3}{y}^{2}\) is continuous at every point \((x,y)\) in the \(xy\text{-plane}\) and \(g(x)=\text{cos}\ x\) is continuous at every real number \(x,\) the continuity of the composition of functions tells us that \(g(x,y)=\text{cos}(4{x}^{3}{y}^{2})\) is continuous at every point \((x,y)\) in the \(xy\text{-plane.}\)

10. Show that the functions \(f(x,y)=2{x}^{2}{y}^{3}+3\) and \(g(x,y)={(2{x}^{2}{y}^{3}+3)}^{4}\) are continuous everywhere.

    जवाफ प्रकट गर्नुहोस्

    The polynomials \(g(x)=2{x}^{2}\) and \(h(y)={y}^{3}\) are continuous at every real number; therefore, by the product of continuous functions theorem, \(f(x,y)=2{x}^{2}{y}^{3}\) is continuous at every point \((x,y)\) in the \(xy\text{-plane.}\) Furthermore, any constant function is continuous everywhere, so \(g(x,y)=3\) is continuous at every point \((x,y)\) in the \(xy\text{-plane.}\) Therefore, \(f(x,y)=2{x}^{2}{y}^{3}+3\) is continuous at every point \((x,y)\) in the \(xy\text{-plane.}\) Last, \(h(x)={x}^{4}\) is continuous at every real number \(x,\) so by the continuity of composite functions theorem \(g(x,y)={(2{x}^{2}{y}^{3}+3)}^{4}\) is continuous at every point \((x,y)\) in the \(xy\text{-plane.}\)

11. Find \(\underset{(x,y,z)\to (4,1,-3)}{\text{lim}}\frac{{x}^{2}y-3z}{2x+5y-z}.\)

    जवाफ प्रकट गर्नुहोस्

    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,

    \[\begin{array}{ll}\underset{(x,y,z)\to (4,1,-3)}{\text{lim}}(2x+5y-z) & =2(\underset{(x,y,z)\to (4,1,-3)}{\text{lim}}x)+5(\underset{(x,y,z)\to (4,1,-3)}{\text{lim}}y)-(\underset{(x,y,z)\to (4,1,-3)}{\text{lim}}z) \\ & =2(4)+5(1)-(-3) \\ & =16.\end{array}\]

    Since this is nonzero, we next find the limit of the numerator. Using the product law, difference law, constant multiple law, and identity law,

    \[\begin{array}{ll}\underset{(x,y,z)\to (4,1,-3)}{\text{lim}}({x}^{2}y-3z) & ={(\underset{(x,y,z)\to (4,1,-3)}{\text{lim}}x)}^{2}(\underset{(x,y,z)\to (4,1,-3)}{\text{lim}}y)-3\underset{(x,y,z)\to (4,1,-3)}{\text{lim}}z \\ & =({4}^{2})(1)-3(-3) \\ & =16+9 \\ & =25.\end{array}\]

    Last, applying the quotient law:

    \[\begin{array}{ll}\underset{(x,y,z)\to (4,1,-3)}{\text{lim}}\frac{{x}^{2}y-3z}{2x+5y-z} & =\frac{\underset{(x,y,z)\to (4,1,-3)}{\text{lim}}({x}^{2}y-3z)}{\underset{(x,y,z)\to (4,1,-3)}{\text{lim}}(2x+5y-z)} \\ & =\frac{25}{16}.\end{array}\]

12. Find \(\underset{(x,y,z)\to (4,-1,3)}{\text{lim}}\sqrt{13-{x}^{2}-2{y}^{2}+{z}^{2}}.\)

    जवाफ प्रकट गर्नुहोस्

    \(\underset{(x,y,z)\to (4,-1,3)}{\text{lim}}\sqrt{13-{x}^{2}-2{y}^{2}+{z}^{2}}=2\)

13. \(\underset{(x,y)\to (1,2)}{\text{lim}}x\)

14. \(\underset{(x,y)\to (1,2)}{\text{lim}}\frac{5{x}^{2}y}{{x}^{2}+{y}^{2}}\)

    जवाफ प्रकट गर्नुहोस्

    2.0

15. Show that the limit \(\underset{(x,y)\to (0,0)}{\text{lim}}\frac{5{x}^{2}y}{{x}^{2}+{y}^{2}}\) exists and is the same along the paths: \(y\text{-axis}\) and \(x\text{-axis,}\) and along \(y=x.\)

16. \(\underset{(x,y)\to (0,0)}{\text{lim}}\frac{4{x}^{2}+10{y}^{2}+4}{4{x}^{2}-10{y}^{2}+6}\)

    जवाफ प्रकट गर्नुहोस्

    \(\frac{2}{3}\)

17. \(\underset{(x,y)\to (11,13)}{\text{lim}}\sqrt{\frac{1}{xy}}\)

18. \(\underset{(x,y)\to (0,1)}{\text{lim}}\frac{{y}^{2}\text{sin}\ x}{x}\)

    जवाफ प्रकट गर्नुहोस्

    \(1\)

19. \(\underset{(x,y)\to (0,0)}{\text{lim}}\text{sin}(\frac{{x}^{8}+{y}^{7}}{x-y+10})\)

20. \(\underset{(x,y)\to (\pi \text{/}4,1)}{\text{lim}}\frac{y\ \text{tan}\ x}{y+1}\)

    जवाफ प्रकट गर्नुहोस्

    \(\frac{1}{2}\)

21. \(\underset{(x,y)\to (0,\pi \text{/}4)}{\text{lim}}\frac{\text{sec}\ x+2}{3x-\text{tan}\ y}\)

22. \(\underset{(x,y)\to (2,5)}{\text{lim}}(\frac{1}{x}-\frac{5}{y})\)

    जवाफ प्रकट गर्नुहोस्

    \(-\frac{1}{2}\)

23. \(\underset{(x,y)\to (4,4)}{\text{lim}}x\ \text{ln}\ y\)

24. \(\underset{(x,y)\to (4,4)}{\text{lim}}{e}^{\text{-}{x}^{2}-{y}^{2}}\)

    जवाफ प्रकट गर्नुहोस्

    \({e}^{-32}\)

25. \(\underset{(x,y)\to (0,0)}{\text{lim}}\sqrt{9-{x}^{2}-{y}^{2}}\)

26. \(\underset{(x,y)\to (1,2)}{\text{lim}}({x}^{2}{y}^{3}-{x}^{3}{y}^{2}+3x+2y)\)

    जवाफ प्रकट गर्नुहोस्

    \(11.0\)

27. \(\underset{(x,y)\to (\pi ,\pi )}{\text{lim}}x\ \text{sin}(\frac{x+y}{4})\)

28. \(\underset{(x,y)\to (0,0)}{\text{lim}}\frac{xy+1}{{x}^{2}+{y}^{2}+1}\)

    जवाफ प्रकट गर्नुहोस्

    \(1.0\)

29. \(\underset{(x,y)\to (0,0)}{\text{lim}}\frac{{x}^{2}+{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}+1}-1}\)

30. \(\underset{(x,y)\to (0,0)}{\text{lim}}\text{ln}({x}^{2}+{y}^{2})\)

    जवाफ प्रकट गर्नुहोस्

    The limit does not exist because when \(x\) and \(y\) both approach zero, the function approaches \(\text{ln}\ 0,\) which is undefined (approaches negative infinity).

31. A point \(({x}_{0},{y}_{0})\) in a plane region \(R\) is an interior point of \(R\) if _________________.

32. A point \(({x}_{0},{y}_{0})\) in a plane region \(R\) is called a boundary point of \(R\) if ___________.

    जवाफ प्रकट गर्नुहोस्

    every open disk centered at \(({x}_{0},{y}_{0})\) contains points inside \(R\) and outside \(R\)

33. \(\underset{(x,y)\to (2,1)}{\text{lim}}\frac{x-y-1}{\sqrt{x-y}-1}\)

34. \(\underset{(x,y)\to (0,0)}{\text{lim}}\frac{{x}^{4}-4{y}^{4}}{{x}^{2}+2{y}^{2}}\)

    जवाफ प्रकट गर्नुहोस्

    \(0.0\)

35. \(\underset{(x,y)\to (0,0)}{\text{lim}}\frac{{x}^{3}-{y}^{3}}{x-y}\)

36. \(\underset{(x,y)\to (0,0)}{\text{lim}}\frac{{x}^{2}-xy}{\sqrt{x}-\sqrt{y}}\)

    जवाफ प्रकट गर्नुहोस्

    \(0.00\)

37. \(\underset{(x,y,z)\to (1,2,3)}{\text{lim}}\frac{x{z}^{2}-{y}^{2}z}{xyz-1}\)

38. \(\underset{(x,y,z)\to (0,0,0)}{\text{lim}}\frac{{x}^{2}-{y}^{2}-{z}^{2}}{{x}^{2}+{y}^{2}-{z}^{2}}\)

    जवाफ प्रकट गर्नुहोस्

    The limit does not exist.

39. \(\underset{(x,y)\to (0,0)}{\text{lim}}\frac{xy+{y}^{3}}{{x}^{2}+{y}^{2}}\)

    1. Along the \(x\text{-axis}\) \((y=0)\)
    2. Along the \(y\text{-axis}\) \((x=0)\)
    3. Along the path \(y=2x\)

40. Evaluate \(\underset{(x,y)\to (0,0)}{\text{lim}}\frac{xy+{y}^{3}}{{x}^{2}+{y}^{2}}\) using the results of previous problem.

    जवाफ प्रकट गर्नुहोस्

    The limit does not exist. The function approaches two different values along different paths.

तपाईँको आफ्नै प्रयास गर्नुहोस्

Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

यसमा थप Multivariable Calculus

Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems