Question

In Exercises \(33-44,\) find the limit \(L\). Then use the \(\varepsilon-\delta\) definition to prove that the limit is \(L\). $$ \lim _{x \rightarrow 2}(x+3) $$

Short answer

The limit \(L\) of the function \(x+3\) as \(x\) approaches 2 is 5, which is confirmed by the \(\varepsilon-\delta\) definition with \(\delta = \varepsilon \).

Step-by-step solution

Find the limit

As \(x\) approaches 2, the function \(x+3\) also approaches the value at 2, which is \(2+3=5\). So, the limit \(L=5\). This can be written as \( \lim _{x \rightarrow 2}(x+3) = 5 \)

Using the epsilon-delta definition

Now the problem moves to proof of the limit using the \(\varepsilon-\delta\) definition. According to this definition, for each \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x - 2| < \delta\), then \( |(x+3) - 5| < \varepsilon\). Solving the inequality \( |x - 2| < \delta\) gives \( -\delta < x - 2 < \delta\) which gives \( 2-\delta < x < 2+\delta\). Substituting these inequalities in the inequality \( |(x+3) - 5| < \varepsilon\) gives \( -\varepsilon < x+3 -5 < \varepsilon\), which is \( -\varepsilon < x - 2 < \varepsilon\) which is true for \(\delta = \varepsilon\). So, here \(\delta = \varepsilon\). So, the \(\varepsilon-\delta\) definition holds, and the limit is proven to be 5.

Concluding the Result

From this, we can see that the limit of \(x+3\) as \(x\) approaches 2 is indeed 5, and the use of the \(\varepsilon-\delta\) definition confirms this.