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Chapter 11: Problem 37

Prove that if \(f\) is a function of two variables that is differentiable at \((a, b),\) then \(f\) is continuous at \((a, b) .\) Hint: Show that $$ \lim _{(\Delta r, \Delta y) \rightarrow(0,0)} f(a+\Delta x, b+\Delta y)=f(a, b) $$

Short Answer

Expert verified
Differentiability implies continuity at a point, thus \( f \) is continuous at \((a,b)\).

Step by step solution

01

Understand the Definition of Differentiability

A function of two variables, \( f(x, y) \), is differentiable at \((a, b)\) if there exist constants \( A \) and \( B \) and a function \( \epsilon(h, k) \) such that \( \epsilon(h, k) \to 0 \) as \( (h, k) \to (0, 0) \), and\[f(a+h, b+k) = f(a, b) + Ah + Bk + \epsilon(h, k) \sqrt{h^2 + k^2}.\]
02

Express Continuity Requirement

A function \( f(x, y) \) is continuous at \((a, b)\) if \[\lim _{(\Delta x, \Delta y) \rightarrow(0,0)} f(a+\Delta x, b+\Delta y) = f(a, b).\]
03

Rewrite the Limit in Terms of Differentiability

Based on differentiability, we have\[f(a + \Delta x, b + \Delta y) = f(a, b) + A\Delta x + B\Delta y + \epsilon(\Delta x, \Delta y)\sqrt{(\Delta x)^2 + (\Delta y)^2}.\]
04

Show the Limit of Additional Terms Goes to Zero

The terms \( A\Delta x \), \( B\Delta y \), and \( \epsilon(\Delta x, \Delta y)\sqrt{(\Delta x)^2 + (\Delta y)^2} \) must go to zero as \( (\Delta x, \Delta y) \to (0,0) \) due to * \( A\Delta x + B\Delta y \) tends to zero since \(\Delta x\) and \(\Delta y\) go to zero. * \( \epsilon(\Delta x, \Delta y)\sqrt{(\Delta x)^2 + (\Delta y)^2} \to 0 \) as \( (\Delta x, \Delta y) \to (0,0) \).
05

Establish the Continuity of the Function

Thus we see:\[\lim_{(\Delta x, \Delta y) \to (0, 0)} [f(a + \Delta x, b + \Delta y) - f(a, b)] = 0\]This implies \[\lim_{(\Delta x, \Delta y) \to (0,0)} f(a + \Delta x, b + \Delta y) = f(a, b).\]Thus, by the definition of continuity, \( f \) is continuous at \((a, b)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiability in Multi-variable Functions
In multivariable calculus, differentiability extends our understanding of how functions change. A function of two variables, say \( f(x, y) \), is differentiable at a point \((a, b)\) if it can be well-approximated by a linear function at this point. This means that small changes in \(x\) and \(y\) can be understood from the linear approximation:
  • The function must be expressible in the form: \[ f(a+h, b+k) = f(a, b) + Ah + Bk + \epsilon(h, k) \sqrt{h^2 + k^2} \]
  • Here, \(A\) and \(B\) are constants that are akin to partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) at \((a, b)\).
  • The term \(\epsilon(h, k)\) goes to zero as the point \((h, k)\) approaches \((0, 0)\), ensuring that \(f(x, y)\) becomes a good linear approximation when close to \((a, b)\).
Differentiability is a stronger condition than continuity and implies that the tangent plane to the surface is a good approximation at that point.
Understanding Two-variable Functions
Functions of two variables are mappings from the plane to the real numbers, expressed as \( f(x, y) \). These functions can represent surfaces in a three-dimensional space.
  • They require both an \(x\) and a \(y\) coordinate, representing positions in a rectangular plane, and they output a single value \(z = f(x, y)\).
  • This value can be visualized as the height of a surface above the \((x, y)\) plane.
  • Common examples include functions such as \( f(x, y) = x^2 + y^2 \), which would produce a parabolic surface.
Two-variable functions often involve examining how the output (or height) changes as one or both of the input variables change. This can involve exploring various characteristics like partial derivatives and second-order derivatives, but understanding these basic elements can make more advanced topics accessible.
Limits and Continuity in Two-variable Functions
Continuity in multivariable functions means that small changes in the input variables \(x\) and \(y\) only cause small changes in the output value \(f(x, y)\). For a function \( f(x, y) \) to be continuous at a point \((a, b)\), it must satisfy the condition:
  • \[ \lim_{(\Delta x, \Delta y) \to (0,0)} f(a+\Delta x, b+\Delta y) = f(a, b) \]
  • This asserts that as the values of \(\Delta x\) and \(\Delta y\) tend to zero, the function value approaches \( f(a, b) \).
  • This lack of sudden jumps or holes in the output makes the function continuous at \((a, b)\).
It's important to note that differentiability implies continuity. If a function is differentiable at any given point, it is necessarily continuous there. Differentiability provides a stricter criterion than continuity, making it a useful property for analyzing and working with functions.

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Most popular questions from this chapter

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=x y+\frac{1}{x}+\frac{1}{y}$$

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