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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]
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