Question

Find the limit \(L .\) Then find \(\delta>0\) such that \(|f(x)-L|<0.01\) whenever \(0<|x-c|<\delta\). $$ \lim _{x \rightarrow 2}\left(x^{2}-3\right) $$

Short answer

The limit \(L\) of the function as \(x\) approaches 2 is 1, and the \(\delta\) that satisfies the conditions of the exercise is approximately 0.005.

Step-by-step solution

Calculate Limit

Calculate the limit of the function as \(x\) approaches 2. Substitution method can be used here. Replace \(x\) with 2 in the equation \(x^{2}-3\) which results in \(2^{2}-3 = 4-3 = 1\). So, the limit \(L\) of the function as \(x\) approaches 2 is 1.

Apply the \(\delta-\varepsilon\) Definition of a Limit

The \(\delta-\varepsilon\) definition of a limit states that for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if \(0<|x-c|<\delta\), then \(|f(x)-L|<\varepsilon\). Here, \(L\) is 1, \(c\) is 2, and \(\varepsilon\) is 0.01. The goal is to find \(\delta\) such that \(|f(x)-1|<0.01\) whenever \(0<|x-2|<\delta\).

Simplifying the Expression

We simplify the expression \(|f(x)-1|<0.01\) to \(|x^2 - 4|<0.01\). This becomes \(-0.01 < x^2 - 4 < 0.01\). So, \(1.99 < x^2 < 2.01\). Taking square roots of these inequalities gives us approximately \(-\delta < x - 2 < \delta\), where \(\delta\) takes the smaller value between \(2 - \sqrt{1.99}\) and \(\sqrt{2.01} - 2\), which is approximately 0.005

Final Result

So, by using the definition of a limit, we found \(\delta \approx 0.005\). Therefore, whenever \(0 < |x - 2| < 0.005\), we have \(|f(x) - 1| < 0.01\).