Question

In Exercises \(101-104,\) use the definition of limits at infinity to prove the limit. $$ \lim _{x \rightarrow-\infty} \frac{1}{x-2}=0 $$

Short answer

Based on the steps followed, it can be concluded that \(\lim_{x \rightarrow -\infty} \frac{1}{x-2} = 0\) is proven by using the definition of limits at infinity.

Step-by-step solution

Define the limit

The task is to prove that for all \(\varepsilon > 0\), there exists a number \(M\) such that if \(x < M\), then \(|f(x) - L| < \varepsilon\), where \(f(x) = \frac{1}{x-2}\), and \(L = 0\). In this context, \(f(x)\) is \( \frac{1}{x-2}\), \(L\) is 0, \(\varepsilon\) is any positive number, and \(M\) will be determined.

Apply the definition

Using the definition, we must show that for any positive number \(\varepsilon\), we can find a sufficiently negative value \(M\), such that if \(x < M\), then \(|\frac{1}{x-2}| < \varepsilon\). We express the absolute value as \(\frac{-1}{x-2}\) when \(x < 2\).

Find the value of M

We solve the inequality \(\frac{-1}{x-2} < \varepsilon\) for \(x\). Manipulating the inequality, we get \(x < \frac{-1}{\varepsilon} + 2 = M\). Hence, for \(x < M\), the absolute inequality holds.

Draw the conclusion

Since for any given \(\varepsilon > 0\), a corresponding \(M\) can be found that fulfills the requirements of the definition of the limit, it is proven that \(\lim_{x \rightarrow -\infty} \frac{1}{x-2} = 0\).