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Chapter 4: Problem 10

Give examples of functions \(f\) and \(g\) such that \(f\) and \(g\) do not have limits at a point \(c\), but such that both \(f+g\) and \(f g\) have limits at \(c\).

Short Answer

Expert verified
The functions \(f(x) = 1/x\) and \(g(x) = -1/x\) satisfy the conditions of the problem. Neither \(f\) nor \(g\) have a limit at \(c=0\), but their sum \(f+g\) and product \(f.g\) both have limits at \(c=0\)(0 and -infinity respectively).

Step by step solution

01

Selection of Functions

Let's choose two functions. A simple choice could be \(f(x) = 1/x\) and \(g(x) = -1/x\).
02

Check Limit of Individual Functions at Point c=0

Both functions \(f\) and \(g\) do not have a limit at \(c=0\). For \(f(x)\), as \(x\) approaches 0, the value of function \(f\) becomes infinite. Likewise, for \(g(x)\), the value becomes negative infinite.
03

Check Limit of Sum and Product of Functions at Point c=0

Now, check the sum and product of the functions at \(c=0\). The sum \(f+g = 1/x - 1/x = 0\), which is well-defined for all x 鈮� 0, and as \(x\) approaches 0, the sum also approaches 0. Hence, limit exists. Similarly, the product \(f.g = (1/x).(-1/x) = -1/x虏\), which is well-defined for all x 鈮� 0, and for \(x\) approaching 0, the product approaches -infinity. Hence, limit exists.
04

Conclusion

Therefore, the functions \(f(x) = 1/x\) and \(g(x) = -1/x\) satisfy the conditions of the problem; they do not have limits at \(c=0\), but their sum and product do have limits at \(c=0\).

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Most popular questions from this chapter

Let \(A \subseteq \mathbb{R}\), let \(f: A \rightarrow \mathbb{R}\) and let \(c \in \mathbb{R}\) be a cluster point of \(A .\) If \(\lim _{\tau \rightarrow c} f\) exists, and if \(|f|\) denotes the function defined for \(x \in A\) by \(|f|(x):=|f(x)|\), prove that \(\lim _{x \rightarrow c}|f|=\left|\lim _{x \rightarrow c} f\right|\)

Let \(I\) be an interval in \(\mathbb{R}\). let \(f: I \rightarrow \mathbb{R}\), and let \(c \in I\). Suppose there exist constants \(K\) and \(L\) such that \(|f(x)-L| \leq K|x-c|\) for \(x \in I .\) Show that \(\lim _{x \rightarrow c} f(x)=L\)

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\), let \(J\) be a closed interval in \(\mathbb{R}\), and let \(c \in J .\) If \(f_{2}\) is the restriction of \(f\) to \(J\), show that if \(f\) has a limit at \(c\) then \(f_{2}\) has a limit at \(c\). Show by example that it does not follow that if \(f_{2}\) has a limit at \(c\), then \(f\) has a limit at \(c\).

Give an example of a function that has a right-hand limit but not a left-band limit at a point.

Determine whether the following limits exist in \(\mathbb{K}\). (a) \(\lim _{x \rightarrow 0} \sin \left(1 / x^{2}\right) \quad(x \neq 0)\) (b) \(\lim _{x \rightarrow 0} x \sin \left(1 / x^{2}\right) \quad(x \neq 0)\) (c) \(\lim _{x \rightarrow 0} \operatorname{sgn} \sin (1 / x) \quad(x \neq 0)\), (d) \(\lim _{x \rightarrow 0} \sqrt{x} \sin \left(1 / x^{2}\right) \quad(x>0)\).
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