Define Multivariable Limits

Multivariable calculus is an extension of single variable calculus. It involves several variables instead of just one. Assume there is an open set containing points (x₀, y₀), let f be a function defined in that open interval except for the points (x₀, y₀). The multivariable limit of this function as it approaches the points (x₀, y₀) and is denoted as:

$\underset{\left(x,y\right)\to (x_0,y_0)}{\lim }f(x,y)=L$

The above statement ϵ−δ using definition means that

$\left|f\left(x,y\right)-L\right|<ϵ whenever 0 < \sqrt{{(x-x_0)}^2+{(y-y_0)}^2}<\delta$

We check for value of the limit as variables x and y approach the points x₀ and y₀.But the approach is different unlike single variable calculus. In multivariable limits, we check for the value of the limit as the function approaches the desired values from different paths.

For a multivariable limit to exist, the function should approach the same value regardless of the path taken to approach the point, which means an infinite number of paths need to be checked. For example:

$f\left(x,y\right)=\underset{\left(x,y\right)\to (0,0)}{\lim } \frac{x^3{y}^3}{3x^6+2y^6}$

If the values for x and y are plugged in the expression, the function returns a value $\frac{0}{0}$. Upon inspection, we can see that the function cannot be factorized as well. Hence, we check the value for this function by approaching it from different paths.

Case 1: Approach along the x-axis.

On the x-axis, the value of the y coordinate is 0. Hence, substitute y = 0 in the expression.

$f\left(x,y\right)=\underset{\left(x,y\right)\to (0,0)}{\lim } \frac{x^3{\left(0\right)}^3}{3x^6+2{\left(0\right)}^6}=0$

Case 2: Approach along the y-axis.

On y-axis, the value of x coordinate is 0. Hence substitute x=0 in the expression.

$f\left(x,y\right)=\underset{\left(x,y\right)\to (0,0)}{\lim } \frac{{\left(0\right)}^3 y^3}{3{\left(0\right)}^6+2y^6}=0$

Case 3: Approach along the line x=y.

$f\left(x,y\right)=\underset{\left(x,y\right)\to (0,0)}{\lim } \frac{x^3 x^3}{3x^6+2x^6}= \underset{\left(x,y\right)\to (0,0)}{\lim } \frac{x^6}{5x^6}=\frac{1}{5}$

As different values are obtained for the limit as different path is taken the multivariable limit does not exist.

Derivatives of Multivariable Functions

A given function may be dependent on more than one variable. For instance, the rate at which a bucket of water is filled depends upon the volume of the bucket, rate of flow of water, cross section of the pipe, etc. The solution to the problem as well as the method of solving varies depending upon the nature of the restrictions. We use multivariable chain rule in such cases to obtain the derivative of multivariable functions and solve the question.

Example: Compute the derivative for the function expressed as $f\left(x,y\right)=x{y}^2$ where

$x\left(t\right)=\cos \left(t\right) and y\left(t\right) = \sin \left(t\right)$

Solution: To compute $\frac{d}{dt}f\left(x\right(t),y(t\left)\right)$, simplify the expression as per the given function.

$$\begin{gathered} \frac{d}{dt}f\left(x\right(t),y(t\left)\right) = \frac{d}{dt}x\left(t\right){\left\{y\left(t\right)\right\}}^2 \\ =\frac{d}{dt}\cos \left(t\right){\sin }^2\left(t\right) \\ =\cos \left(t\right)2\sin \left(t\right)\cos \left(t\right)-\sin \left(t\right){\sin }^2\left(t\right) \\ =2\sin \left(t\right){\cos }^2\left(t\right)-{\sin }^3\left(t\right) \end{gathered}$$

Partial Derivatives of Multivariable Functions

We make use of the concept of partial derivatives to find the derivative of a multivariable function with an n-variable input. Partial derivatives are used to measure rates of change, approximating formulas and identifying points of local maxima and minima.

Unlike ordinary derivatives, partial derivatives measure the rate of change, but with respect to one variable. Multivariable partial derivatives are the rates of change with respect to each variable separately. Every other variable is treated as a constant. The example listed above can also be solved using the partial derivatives in the form of a chain rule.

In mathematical form:

$\frac{d}{dt}f\left(x\right(t),y(t\left)\right)=\frac{\partial f}{\partial x}·\frac{dx}{dt}+\frac{\partial f}{\partial y}·\frac{dy}{dt}$

To solve the example shown above using chain rule, compute the partial derivative with respect to x and y.

Treating y as a constant, we get:

$\frac{\partial f}{\partial x}=y^2$

Similarly, treating x as a constant, we get:

$\frac{\partial f}{\partial y}=2xy$

Plug the values in the expression given for chain rule.

$$\begin{gathered} \frac{\partial f}{\partial x}·\frac{dx}{dt}+\frac{\partial f}{\partial y}·\frac{dy}{dt}= y^2\left(t\right)·\frac{d}{dt}x\left(t\right)+2x\left(t\right)y\left(t\right)·\frac{d}{dt}y\left(t\right) \\ =y^2\left(t\right)·\frac{d}{dt}\cos \left(t\right)+2x\left(t\right)y\left(t\right)·\frac{d}{dt}\sin \left(t\right) \\ =y^2\left(t\right)·\left\{-\sin \left(t\right)\right\}+2x\left(t\right)y\left(t\right)·\left\{\cos \left(t\right)\right\} \\ ={\sin }^2\left(t\right)·\left\{-\sin \left(t\right)\right\}+2\sin \left(t\right)\cos \left(t\right)·\left\{\cos \left(t\right)\right\} \\ =-{\sin }^3\left(t\right)+2\sin \left(t\right){\cos }^2\left(t\right) \end{gathered}$$

The result with partial derivatives is the same as that obtained with regular derivative.

The ordinary derivative of a composition of a multivariable function is equal summation of all the partial derivatives with respect to the multivariable multiplied by derivative of that multivariable with respect to the intermediate variable.

Implicit Differentiation to Find Partial Derivatives

Consider the equation given by:

$$\begin{gathered} x^3+y^3 = y·\sin x \\ \Rightarrow x^3+y^3-y·\sin x = 0 \end{gathered}$$

A multivariable function in terms of x and y can be written as:

$F(x,y)=x^3+y^3-y·\sin x$

Obtain the partial derivative with respect to x:

$\frac{\partial F}{\partial x}=3x^2-y \cos x$

Obtain the partial derivative with respect to y:

$\frac{\partial F}{\partial y}=3y^2+\sin x$

In order to calculate $\frac{dy}{dx}$, use the formula $\frac{dy}{dx}=-\frac{{\displaystyle \frac{\partial F}{\partial x}}}{{\displaystyle \frac{\partial F}{\partial y}}}$

$\Rightarrow -\frac{{\displaystyle \raisebox{1ex}{$\partial F$}\!\left/ \!\raisebox{-1ex}{$\partial x$}\right.}}{{\displaystyle \raisebox{1ex}{$\partial F$}\!\left/ \!\raisebox{-1ex}{$\partial y$}\right.}}=-\left(\frac{3x^2-y \cos x}{3y^2+\sin x}\right)=\frac{y \cos x-3x^2}{3y^2 + \sin x }$

Squeeze Theorem

In cases where the limit of a function which cannot be evaluated at a given point, we make use of the squeeze theorem. To apply squeeze theorem to a function f(x), consider two functions on the lower and upper boundary of f(x). That is g(x) and h(x) respectively.

Using inequalities, we can express the relation between the three functions as:

$g\left(x\right)\le f\left(x\right)\le h\left(x\right)$

If the value for f(x) is to be evaluated as x approaches a value a, then it must satisfy the following criteria for a positive number N

$0<\left|x-a\right|<N$

If the criteria are satisfied, we can that

$\underset{x\to a}{\lim }g\left(x\right) = \underset{x\to a}{\lim }h\left(x\right) = L$

By squeeze theorem, it can be concluded that

$\underset{x\to a}{\lim }f\left(x\right) =L$

Related Solutions on bartleby

https://www.bartleby.com/questions-and-answers/evaluate-the-limit-if-it-exists-or-show-that-the-limit-does-not-exist.-provide-justification.-cos2xy/245f3db1-07c4-46ce-a181-c4fc2effa138

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