1. We are asked to compute limits using the (\epsilon - \delta) definition of limits for the following functions: (a) $\lim_{x \to 1} (x^2 + 1)$ (b) $\lim_{x \to c} x$ where $c \neq 0$ (c) $\lim_{x \to -2} \csc x$ (d) $\lim_{x \to 0} \frac{3x^2 + 7x + 2}{2x + 4}$ --- 2. Recall the (\epsilon - \delta) definition of limit: For a function $f(x)$, $\lim_{x \to a} f(x) = L$ means: For every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \epsilon$. This means we can make $f(x)$ as close as we want to $L$ by choosing $x$ sufficiently close to $a$. --- 3. (a) Compute $\lim_{x \to 1} (x^2 + 1)$: - The function is $f(x) = x^2 + 1$. - We expect the limit to be $f(1) = 1^2 + 1 = 2$. - To prove using (\epsilon - \delta): We want $|x^2 + 1 - 2| = |x^2 - 1| < \epsilon$. Factor: $|x^2 - 1| = |x - 1||x + 1|$. Choose $\delta \leq 1$ so that if $|x - 1| < \delta$, then $x$ is in $(0, 2)$, so $|x + 1| < 3$. Then $|x^2 - 1| < 3|x - 1| < \epsilon$ if $|x - 1| < \frac{\epsilon}{3}$. - So choose $\delta = \min(1, \frac{\epsilon}{3})$. - Hence, $\lim_{x \to 1} (x^2 + 1) = 2$. --- 4. (b) Compute $\lim_{x \to c} x$ where $c \neq 0$: - The function is $f(x) = x$. - The limit is clearly $c$. - Using (\epsilon - \delta): We want $|x - c| < \epsilon$. - Choose $\delta = \epsilon$. - Then if $|x - c| < \delta$, $|f(x) - c| = |x - c| < \epsilon$. - So $\lim_{x \to c} x = c$. --- 5. (c) Compute $\lim_{x \to -2} \csc x$: - $\csc x = \frac{1}{\sin x}$. - Since $\sin(-2) \neq 0$, $\csc x$ is continuous at $x = -2$. - So $\lim_{x \to -2} \csc x = \csc(-2) = \frac{1}{\sin(-2)}$. - Using (\epsilon - \delta) definition, continuity of $\sin x$ and reciprocal function applies. --- 6. (d) Compute $\lim_{x \to 0} \frac{3x^2 + 7x + 2}{2x + 4}$: - Evaluate at $x=0$: Numerator: $3(0)^2 + 7(0) + 2 = 2$ Denominator: $2(0) + 4 = 4$ - So limit is $\frac{2}{4} = \frac{1}{2}$. - Using (\epsilon - \delta): We want $\left| \frac{3x^2 + 7x + 2}{2x + 4} - \frac{1}{2} \right| < \epsilon$. - Simplify the difference: $$\left| \frac{3x^2 + 7x + 2}{2x + 4} - \frac{1}{2} \right| = \left| \frac{2(3x^2 + 7x + 2) - (2x + 4)}{2(2x + 4)} \right| = \left| \frac{6x^2 + 14x + 4 - 2x - 4}{2(2x + 4)} \right| = \left| \frac{6x^2 + 12x}{2(2x + 4)} \right|$$ $$= \left| \frac{6x^2 + 12x}{4x + 8} \right| = \left| \frac{6x(x + 2)}{4(x + 2)} \right| = \left| \frac{6x}{4} \right| = \frac{3}{2} |x|$$ - To make this less than $\epsilon$, choose $|x| < \frac{2}{3} \epsilon$. - Also ensure denominator $2x + 4 \neq 0$ by restricting $x$ near 0. - So choose $\delta = \min(1, \frac{2}{3} \epsilon)$. - Hence, $\lim_{x \to 0} \frac{3x^2 + 7x + 2}{2x + 4} = \frac{1}{2}$. --- Final answers: (a) 2 (b) c (c) $\csc(-2)$ (d) $\frac{1}{2}$