Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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]
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]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.