Problem 5 Explain how examining limits alo... [FREE SOLUTION]

Chapter 13: Problem 5

Explain how examining limits along multiple paths may prove the nonexistence of a limit.

Short Answer

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Answer: Examining limits along multiple paths helps prove the nonexistence of a limit because if a limit exists, it must have the same and unique value regardless of which path is taken to approach the point. By analyzing different paths and finding different values for the limit or finding that the limit does not converge along some paths, we can conclude that the limit does not exist for the given function. This approach is important in determining the existence and uniqueness of a limit for a given function.

Step by step solution

Definition of Limit

In calculus, the limit of a function at a particular point is the value that the function approaches as its input variable approaches that point. Mathematically, it can be denoted as: lim_(x->a) f(x) = L If such a limit exists, it must be true regardless of which path is taken to approach the point `x=a`.

Example: Evaluating a Limit along Different Paths

Let's consider the following function: f(x, y) = (x^2 * y) / (x^4 + y^2) The goal is to find the limit as `(x, y)` approaches `(0, 0)`. We will examine the limit along different paths to determine if the limit exists.

Path 1: y = 0

First, let's take the path when y = 0: lim_(x->0, y->0) f(x, 0) = lim_(x->0) (x^2 * 0) / (x^4 + 0^2) = lim_(x->0) 0 = 0 As `(x, y)` approaches `(0, 0)` along this path, the limit is found to be 0.

Path 2: x = y

Let's now take another path, when x = y: lim_(x->0, y->0) f(x, x) = lim_(x->0) (x^2 * x) / (x^4 + x^2) = lim_(x->0) (x^3) / (x^2 * (x^2 + 1)) Canceling out `x^2` from the numerator and the denominator, we get: lim_(x->0) x / (x^2 + 1) = 0 / (0^2 + 1) = 0 As `(x, y)` approaches `(0, 0)` along this path, the limit is found to be 0.

Path 3: y = x^2

Let's try a different path, when y = x^2: lim_(x->0, y->0) f(x, x^2) = lim_(x->0) (x^2 * x^2) / (x^4 + (x^2)^2) = lim_(x->0) (x^4) / (x^4 * (1 + x^2)) Canceling out `x^4` from the numerator and the denominator, we get: lim_(x->0) 1 / (1 + x^2) = 1 / (1 + 0^2) = 1 As `(x, y)` approaches `(0, 0)` along this path, the limit is found to be 1.

Interpreting Results and Conclusion

From the three different paths we've analyzed, we have found different values for the limit when `(x, y)` approaches `(0, 0)`. Since the limit should be the same and unique for all paths, we can conclude that the limit does not exist at `(0, 0)` for the given function. By examining the limits along multiple paths, we have proven the nonexistence of a limit for this particular example. This illustrates the importance of analyzing different paths when determining the existence of a limit for a given function.

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Explain how examining limits along multiple paths may prove the nonexistence of a limit.

Short Answer

Step by step solution

Definition of Limit

Example: Evaluating a Limit along Different Paths

Path 1: y = 0

Path 2: x = y

Path 3: y = x^2

Interpreting Results and Conclusion

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