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Chapter 5: Problem 17

(a) Prove that \(\lim _{x \rightarrow 0} 1 / x\) does not exist, i.e., show that \(\lim _{x \rightarrow 0} 1 / x=l\) is false for every number \(l\) (b) Prove that \(\lim _{x \rightarrow 1} 1 /(x-1)\) does not exist.

Short Answer

Expert verified
(a) The limit \(\lim_{x \to 0} \frac{1}{x}\) does not exist because the left and right limits differ. (b) The limit \(\lim_{x \to 1} \frac{1}{x-1}\) does not exist because the left and right limits differ.

Step by step solution

01

Understanding the limit definition

A limit \(\lim_{x \to a} f(x) = L\) exists if and only if for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < |x - a| < \delta\), then \|f(x) - L| < \epsilon|\. If we can show that this condition is violated for all possible values of \L\, then we can conclude that the limit does not exist.
02

Proving the limit \(\lim_{x \to 0} \frac{1}{x} \) does not exist

Consider the behavior of \(\frac{1}{x}\) as \(x \to 0^+\) and \(x \to 0^-\). When x approaches 0 from the positive side (denoted \(0^+\)), \(\frac{1}{x}\) becomes very large and positive. Conversely, as x approaches 0 from the negative side (denoted \(0^-\)), \(\frac{1}{x}\) becomes very large and negative. Since the limits from the left and the right are not equal (one approaches \infty, while the other approaches -\infty), the overall limit does not exist.
03

Contradiction proof for (a)

Assume for the sake of contradiction that \(\lim_{x \to 0} \frac{1}{x} = l\) for some finite number \l\. Then there must exist a \(\delta > 0\) such that whenever \(0 < |x| < \delta\), \|\frac{1}{x} - l| < \epsilon|\. However, choosing \ x \, very close to zero, we can make \(\frac{1}{x}\) arbitrarily large in magnitude, contradicting the assumption of a finite limit \l\. Thus, \(\lim_{x \to 0} \frac{1}{x}\) does not exist.
04

Analyzing \ \lim_{x \to 1} \frac{1}{x-1} \

Now consider the behavior of \(\frac{1}{x-1}\) as \(x \to 1^+\) and \(x \to 1^-\). When x approaches 1 from the right side (denoted \(1^+\)), \(\frac{1}{x-1}\) becomes very large and positive. As x approaches 1 from the left side (denoted \(1^-\)), \(\frac{1}{x-1}\) becomes very large and negative. Since the limits from the left and the right are not equal (one approaches \infty\, while the other approaches -infinity), the limit \(\lim_{x \to 1} \frac{1}{x-1}\) does not exist.
05

Contradiction proof for (b)

Assume for the sake of contradiction that \(\lim_{x \to 1} \frac{1}{x-1} = k\) for some finite number \k\. Then there must exist a \(\delta > 0\) such that whenever \(0 < |x - 1| < \delta\), \|\frac{1}{x-1} - k| < \epsilon|\. However, choosing \ x \ very close to 1, we can make \(\frac{1}{x-1}\) arbitrarily large in magnitude, contradicting the assumption of a finite limit \k\. Thus, \(\lim_{x \to 1} \frac{1}{x-1}\) does not exist.

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Key Concepts

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

Limit Definition
In calculus, the concept of a limit is fundamental. It's the value that a function approaches as the input approaches some value. Formally, for a function f(x) and a point a, the limit \(\lim_{{x \to a}} f(x) = L\) means that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < |x - a| < \delta\), it follows that \(|f(x) - L| < \epsilon\). This precise definition is crucial for proving whether a limit exists or not. To paraphrase, the function should get arbitrarily close to L as x gets close to a. If we cannot find such a delta for every epsilon, the limit does not exist. Understanding this definition helps in recognizing and proving limits.
Epsilon-Delta Method
The epsilon-delta method provides a rigorous approach to proving limits. This method involves two parts: epsilon (\(\epsilon\)) representing how close we want \(f(x)\) to be to the limit L, and delta (\(\delta\)) representing how close x needs to be to a. To illustrate, let's consider the limit \( \lim_{{x \to 0}} \frac{1}{x} \):
The behavior of \( \frac{1}{x} \) as x approaches 0 is key.
  • If x approaches 0 from the positive side (\(0^+\)), \( \frac{1}{x} \) becomes very large and positive.
  • If x approaches 0 from the negative side (\(0^-\)), \( \frac{1}{x} \) becomes very large and negative.
Since the left-hand limit (approaching from the negative side) and the right-hand limit (approaching from the positive side) are not the same, we use the epsilon-delta method to show there's no single limit L that satisfies the definition. This contradiction proves that \( \lim_{{x \to 0}} \frac{1}{x} \) does not exist. The epsilon-delta method is particularly useful when the behavior of a function is not immediately obvious or intuitive.
Behavior at Infinity
Understanding the behavior of a function at infinity helps in verifying limits that involve extreme values. For instance, consider \( \frac{1}{x} \) as x approaches 0. To explore the behavior:
  • When x approaches 0 from the positive side (\(0^+\)), \( \frac{1}{x} \rightarrow \infty \).
  • When x approaches 0 from the negative side (\(0^-\)), \( \frac{1}{x} \rightarrow -\infty \).
Since moving towards infinity or negative infinity shows drastically different outcomes, their absolute magnitudes increase without bound but with opposite signs. Similarly, for \( \frac{1}{x-1} \) as x approaches 1, one side trends towards positive infinity while the other trends towards negative infinity. These extreme behaviors indicate that the function does not settle to any finite value, confirming that the limits \( \lim_{{x \to 0}} \frac{1}{x} \) and \( \lim_{{x \to 1}} \frac{1}{x-1} \) do not exist. By examining how functions behave as variables approach infinity, we can draw vital conclusions about their limits or the lack thereof.

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Most popular questions from this chapter

Prove that if \(f(x)=x\) for rational \(x,\) and \(f(x)=-x\) for irrational \(x,\) then \(\lim _{x \rightarrow a} f(x)\) does not exist if \(a \neq 0\)

(a) Prove that \(\lim _{x \rightarrow a} f(x)=l\) if and only if \(\lim _{x \rightarrow a}[f(x)-l]=0 .\) (First see why the assertion is obvious; then provide a rigorous proof. In this chapter most problems which ask for proofs should be treated in the same way.) (b) Prove that \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow a} f(x-a)\) (c) Prove that \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} f\left(x^{3}\right)\) (d) Give an example where \(\lim _{x \rightarrow 0} f\left(x^{2}\right)\) exists, but \(\lim _{x \rightarrow 0} f(x)\) does not.

(a) Define \(\lim _{t \rightarrow a^{+}} f(x)=\infty\) and \(\lim f(x)=\infty .\) (Or at least convince yourself that you could write down the definitions if you had the energy. How many other such symbols can you define? (b) Prove that \(\lim _{x \rightarrow 0^{+}} 1 / x=\infty\) (c) Prove that \(\lim _{x \rightarrow 0^{+}} f(x)=\infty\) if and only if \(\lim _{x \rightarrow \infty} f(1 / x)=\infty\)

Prove that if \(f(x)=0\) for irrational \(x\) and \(f(x)=1\) for rational \(x\) then \(\lim f(x)\) does not exist for any \(a\)

Suppose the functions \(f\) and \(g\) have the following property: for all \(\varepsilon>0\) and all \(x\) $$ \text { if } 0<|x-2|<\sin ^{2}\left(\frac{\varepsilon^{2}}{9}\right)+\varepsilon, \text { then }|f(x)-2|<\varepsilon $$ if \(0<|x-2|<\varepsilon^{2},\) then \(|g(x)-4|<\varepsilon\) For each \(\varepsilon>0\) find a \(\delta>0\) such that, for all \(x\) (i) if \(0<|x-2|<\delta,\) then \(|f(x)+g(x)-6|<\varepsilon\) (ii) if \(0<|x-2|<\delta,\) then \(|f(x) g(x)-8|<\varepsilon\) (iii) if \(0<|x-2|<\delta,\) then \(\left|\frac{1}{g(x)}-\frac{1}{4}\right|<\varepsilon\) (iv) if \(0<|x-2|<\delta,\) then \(\left|\frac{f(x)}{g(x)}-\frac{1}{2}\right|<\varepsilon\)
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