5. By using \( (\varepsilon, \delta) \) definition of the limit, show that \[ \lim _{x \rightarrow-2} x^{2}-1=3, / \lim _{x \rightarrow-2} \frac{x^{2}+x+1}{1-x}=1, \lim _{x \rightarrow 0^{+}} \frac{1}{x}=+\infty, \quad \lim _{x \rightarrow a} \sqrt{x}=\sqrt{a}, a \geq 0 . \]

To demonstrate the given limits using the \((\varepsilon, \delta)\) definition of a limit, we'll address each limit individually. The \((\varepsilon, \delta)\) definition states: > **Definition:** > \(\lim_{x \to c} f(x) = L\) > means that for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < |x - c| < \delta\), it follows that \(|f(x) - L| < \varepsilon\). Let's apply this definition to each limit. --- ### 1. \(\lim_{x \to -2} (x^2 - 1) = 3\) **Goal:** For every \(\varepsilon > 0\), find \(\delta > 0\) such that if \(|x + 2| < \delta\), then \(|x^2 - 1 - 3| = |x^2 - 4| < \varepsilon\). **Proof:** 1. **Express \(|x^2 - 4|\):** \[ |x^2 - 4| = |x - 2||x + 2| \] 2. **Bound \(|x + 2|\):** - Assume \(\delta \leq 1\). Then, if \(|x + 2| < \delta \leq 1\), \(x\) is in the interval \((-3, -1)\). - Therefore, \(|x - 2| = |x + 2 - 4| \leq |x + 2| + 4 < 1 + 4 = 5\). 3. **Combine the bounds:** \[ |x^2 - 4| = |x - 2||x + 2| < 5|x + 2| \] 4. **Set the inequality:** \[ 5|x + 2| < \varepsilon \quad \Rightarrow \quad |x + 2| < \frac{\varepsilon}{5} \] 5. **Choose \(\delta\):** \[ \delta = \min\left\{1, \frac{\varepsilon}{5}\right\} \] **Conclusion:** For this choice of \(\delta\), whenever \(|x + 2| < \delta\), it ensures \(|x^2 - 4| < \varepsilon\), thereby proving that \(\lim_{x \to -2} (x^2 - 1) = 3\). --- ### 2. \(\lim_{x \to -2} \frac{x^2 + x + 1}{1 - x} = 1\) **Goal:** For every \(\varepsilon > 0\), find \(\delta > 0\) such that if \(|x + 2| < \delta\), then \(\left|\frac{x^2 + x + 1}{1 - x} - 1\right| < \varepsilon\). **Proof:** 1. **Simplify the expression:** \[ \left|\frac{x^2 + x + 1}{1 - x} - 1\right| = \left|\frac{x^2 + x + 1 - (1 - x)}{1 - x}\right| = \left|\frac{x^2 + 2x}{1 - x}\right| = \frac{|x(x + 2)|}{|1 - x|} \] 2. **Bound \(|1 - x|\):** - Assume \(\delta \leq 1\). Then, if \(|x + 2| < 1\), \(x \in (-3, -1)\). - In this interval, \(1 - x > 1 - (-3) = 4\), so \(|1 - x| \geq 4\). 3. **Express the inequality:** \[ \frac{|x||x + 2|}{|1 - x|} \leq \frac{|x||x + 2|}{4} \] 4. **Bound \(|x|\):** - From \(x \in (-3, -1)\), \(|x| < 3\). 5. **Combine the bounds:** \[ \frac{|x||x + 2|}{4} < \frac{3|x + 2|}{4} \] 6. **Set the inequality:** \[ \frac{3|x + 2|}{4} < \varepsilon \quad \Rightarrow \quad |x + 2| < \frac{4\varepsilon}{3} \] 7. **Choose \(\delta\):** \[ \delta = \min\left\{1, \frac{4\varepsilon}{3}\right\} \] **Conclusion:** With this \(\delta\), \(\left|\frac{x^2 + x + 1}{1 - x} - 1\right| < \varepsilon\) whenever \(|x + 2| < \delta\), establishing that \(\lim_{x \to -2} \frac{x^2 + x + 1}{1 - x} = 1\). --- ### 3. \(\lim_{x \to 0^+} \frac{1}{x} = +\infty\) **Goal:** For every \(M > 0\), find \(\delta > 0\) such that if \(0 < x < \delta\), then \(\frac{1}{x} > M\). **Proof:** 1. **Set the inequality:** \[ \frac{1}{x} > M \quad \Rightarrow \quad x < \frac{1}{M} \] 2. **Choose \(\delta\):** \[ \delta = \frac{1}{M} \] **Conclusion:** For any \(M > 0\), choosing \(\delta = \frac{1}{M}\) ensures that if \(0 < x < \delta\), then \(\frac{1}{x} > M\). This proves that \(\lim_{x \to 0^+} \frac{1}{x} = +\infty\). --- ### 4. \(\lim_{x \to a} \sqrt{x} = \sqrt{a}\), where \(a \geq 0\) **Goal:** For every \(\varepsilon > 0\), find \(\delta > 0\) such that if \(|x - a| < \delta\) and \(x \geq 0\), then \(|\sqrt{x} - \sqrt{a}| < \varepsilon\). **Proof:** 1. **Express \(|\sqrt{x} - \sqrt{a}|\):** \[ |\sqrt{x} - \sqrt{a}| = \frac{|x - a|}{|\sqrt{x} + \sqrt{a}|} \] 2. **Bound \(|\sqrt{x} + \sqrt{a}|\):** - Assume \(\delta \leq a + 1\) (if \(a = 0\), adjust accordingly). - For \(x\) near \(a\), \(\sqrt{x} + \sqrt{a} \geq \sqrt{0} + \sqrt{a} = \sqrt{a}\). - If \(a > 0\), then \( \sqrt{x} + \sqrt{a} \geq \sqrt{a} \). - If \(a = 0\), \(|\sqrt{x} + \sqrt{a}| = \sqrt{x} \geq \sqrt{x}\), and we handle this case separately. 3. **Set the inequality for \(a > 0\):** \[ \frac{|x - a|}{\sqrt{a}} < \varepsilon \quad \Rightarrow \quad |x - a| < \sqrt{a} \cdot \varepsilon \] 4. **Choose \(\delta\):** \[ \delta = \min\left\{1, \sqrt{a} \cdot \varepsilon\right\} \] 5. **Special Case (\(a = 0\)):** - Here, \(|\sqrt{x} - 0| = \sqrt{x}\). - To have \(\sqrt{x} < \varepsilon\), set \(x < \varepsilon^2\). - Choose \(\delta = \varepsilon^2\). **Conclusion:** For \(a > 0\), choose \(\delta = \min\{1, \sqrt{a} \cdot \varepsilon\}\) to ensure \(|\sqrt{x} - \sqrt{a}| < \varepsilon\) whenever \(|x - a| < \delta\). For \(a = 0\), choose \(\delta = \varepsilon^2\). This establishes that \(\lim_{x \to a} \sqrt{x} = \sqrt{a}\) for \(a \geq 0\). --- Each of these proofs utilizes the \((\varepsilon, \delta)\) framework to rigorously establish the respective limits.