Question

Find the limit \(L\). Then use the \(\varepsilon-\delta\) definition to prove that the limit is \(L\). $$ \lim _{x \rightarrow 4} \sqrt{x} $$

Short answer

The limit \( L \) of the function \( \sqrt{x} \) as \( x \) approaches 4 is 2, as proven by the \( \varepsilon-\delta \) definition of limit.

Step-by-step solution

Finding the limit

First, let's find the limit of \( \sqrt{x} \) as \( x \) approaches 4. This can be done by simply substituting 4 into the function: \( \sqrt{4} = 2 \). So, the limit \( L \) is 2.

Set up \( \varepsilon-\delta \) definition

Next, apply the \( \varepsilon-\delta \) definition. It says that, for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x-4| < \delta \), then \( | \sqrt{x} - 2 | < \varepsilon \). This is what we need to prove.

Manipulating the inequality

Manipulating the inequality \( | \sqrt{x} - 2 | < \varepsilon \), we get \( |x - 4| < ( \varepsilon )^2 \). So, we want to look for a \( \delta \) such that whenever \( 0 < |x - 4| < \delta \), \( |x - 4| < ( \varepsilon )^2 \) holds true.

Choosing the value of \( \delta \)

To fulfill the above stated inequality condition, choose \( \delta = \varepsilon^2 \). Then, for \( 0 < |x - 4| < \varepsilon^2 \) which is essentially \( 0 < |x - 4| < \delta \), we can say that \( |x - 4| < (\varepsilon)^2 \).

Conclusion

Finally, it is now shown that for every \( \varepsilon > 0 \), there exists a \( \delta = \varepsilon^2 \) such that if \( 0 < |x - 4| < \delta \), then \( | \sqrt{x} - 2 | < \varepsilon \). This means the limit of \( \sqrt{x} \) as \( x \) approaches 4 is indeed 2. This statement confirms our initial calculation of the limit was correct.