At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

Prove that [tex]\(\lim_{x \to 0} \sqrt{4 - x} = 2\)[/tex].

Sagot :

To prove that [tex]\(\lim_{{x \to 0}} \sqrt{4-x} = 2\)[/tex], we will use the definition of a limit. Let’s apply the [tex]\(\epsilon\)[/tex]-[tex]\(\delta\)[/tex] definition of a limit:

1. Definition: We need to show that for every [tex]\(\epsilon > 0\)[/tex], there exists a [tex]\(\delta > 0\)[/tex] such that if [tex]\(0 < |x - 0| < \delta\)[/tex], then [tex]\(|\sqrt{4 - x} - 2| < \epsilon\)[/tex].

2. Simplifying the expression: Let's start by manipulating the expression [tex]\(|\sqrt{4 - x} - 2|\)[/tex]. Notice:
[tex]\[ |\sqrt{4 - x} - 2|. \][/tex]

3. Converting to a helpful form: We can multiply and divide by the conjugate to help simplify:
[tex]\[ |\sqrt{4 - x} - 2| = \left|\frac{(\sqrt{4 - x} - 2) \cdot (\sqrt{4 - x} + 2)}{\sqrt{4 - x} + 2}\right| = \left|\frac{4 - x - 4}{\sqrt{4 - x} + 2}\right| = \left|\frac{-x}{\sqrt{4 - x} + 2}\right| = \frac{|x|}{|\sqrt{4 - x} + 2|}. \][/tex]

4. Estimating the denominator: Observe that since [tex]\(x\)[/tex] is approaching 0, [tex]\(\sqrt{4 - x} + 2\)[/tex] is close to [tex]\(\sqrt{4} + 2 = 4\)[/tex]. Let's note that for small values of [tex]\(x\)[/tex], the expression [tex]\(\sqrt{4 - x} + 2\)[/tex] remains close to 4 and is bounded below by some positive number. Specifically, since [tex]\(\sqrt{4 - x}\)[/tex] is a continuous function near [tex]\(x = 0\)[/tex]:

[tex]\[ \sqrt{4 - x} \in [\sqrt{3.9}, 2] \text{ when } x \text{ is close to } 0. \][/tex]

Hence, we can estimate:
[tex]\[ \sqrt{4 - x} + 2 \geq \sqrt{3.9} + 2 > 2 + 2 = 4. \][/tex]

Now, we have a clear lower bound for the denominator which does not equal zero and remains positive.

5. Finding [tex]\(\delta\)[/tex] in terms of [tex]\(\epsilon\)[/tex]:
Given that for very small [tex]\(x\)[/tex], [tex]\(\sqrt{4 - x} + 2\)[/tex] is approximately 4:
[tex]\[ \frac{|x|}{|\sqrt{4 - x} + 2|} \leq \frac{|x|}{4}. \][/tex]

For [tex]\(|\sqrt{4 - x} - 2| < \epsilon\)[/tex], we need:
[tex]\[ \frac{|x|}{4} < \epsilon \implies |x| < 4\epsilon. \][/tex]

Thus, we can take [tex]\(\delta = 4\epsilon\)[/tex].

6. Conclusion: Finally, by the [tex]\(\epsilon\)[/tex]-[tex]\(\delta\)[/tex] definition, for every [tex]\(\epsilon > 0\)[/tex], if we choose [tex]\(\delta = 4\epsilon\)[/tex], then for all [tex]\(x\)[/tex] such that [tex]\(0 < |x - 0| < \delta\)[/tex], it follows that:
[tex]\[ |\sqrt{4 - x} - 2| < \epsilon. \][/tex]

Therefore, we have shown via the [tex]\(\epsilon\)[/tex]-[tex]\(\delta\)[/tex] definition that:
[tex]\[ \lim_{{x \to 0}} \sqrt{4 - x} = 2. \][/tex]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.