8. Une the jrocine (formal defition) of the linit to peove that 8.1. $$ \lim _{x \rightarrow 1}(2 x-3)=3 \quad $$ 8.2. $$ \lim _{x \rightarrow-2}(2 x-3)=-9 \quad 8.3 \lim _{x \rightarrow 1} x^{3}=4 $$

Question

UpStudy AI · Accepted Answer

**8.1. Prove that $$\displaystyle \lim_{x \to 1}(2x-3) = -1$$** Using the epsilon–delta definition of limit, we must show that for every $$\epsilon > 0$$ there exists a $$\delta > 0$$ such that if $$ 0 < |x-1| < \delta \quad \text{then} \quad |(2x-3)-(-1)| < \epsilon. $$ Step 1. Compute the difference: $$ |(2x-3)-(-1)| = |2x-3+1| = |2x-2| = 2|x-1|. $$ Step 2. To ensure $$2|x-1|<\epsilon$$ it is enough to have $$ |x-1| < \frac{\epsilon}{2}. $$ Step 3. Choose $$\delta = \frac{\epsilon}{2}$$. Then, whenever $$ 0 < |x-1| < \delta, $$ we have $$ |(2x-3)-(-1)| = 2|x-1| < 2\delta = \epsilon. $$ Thus, by the formal definition of limit, $$\displaystyle \lim_{x \to 1}(2x-3) = -1$$. --- **8.2. Prove that $$\displaystyle \lim_{x \to -2}(2x-3) = -7$$** For the function $$f(x)=2x-3$$, note that $$f(-2)=2(-2)-3=-7$$. We wish to show that for every $$\epsilon > 0$$ there exists a $$\delta > 0$$ such that if $$ 0 < |x-(-2)| < \delta \quad \text{then} \quad |(2x-3)-(-7)| < \epsilon. $$ Step 1. Compute the difference: $$ |(2x-3)-(-7)| = |2x-3+7| = |2x+4| = 2|x+2|. $$ Step 2. We wish to have $$2|x+2|<\epsilon$$. This will be achieved if $$ |x+2| < \frac{\epsilon}{2}. $$ Step 3. Choose $$\delta = \frac{\epsilon}{2}$$. Then, whenever $$ 0 < |x+2| < \delta, $$ it follows that $$ |(2x-3)-(-7)| = 2|x+2| < 2\delta = \epsilon. $$ Thus, by the formal definition of limit, $$\displaystyle \lim_{x \to -2}(2x-3) = -7$$. --- **8.3. Prove that $$\displaystyle \lim_{x \to 1}x^3 = 1$$** For the function $$f(x)=x^3$$, we note that $$f(1)=1$$. We must show that for every $$\epsilon > 0$$ there exists a $$\delta > 0$$ such that if $$ 0 < |x-1| < \delta \quad \text{then} \quad |x^3-1| < \epsilon. $$ Step 1. Factor the difference: $$ |x^3-1| = |x-1|\;|x^2+x+1|. $$ Step 2. To control the factor $$|x^2+x+1|$$, restrict $$x$$ such that $$|x-1|<1$$. Then $$x \in (0,2)$$. For $$x$$ in this interval, we have $$ x^2 \le 4,\quad x \le 2,\quad \text{so} \quad x^2+x+1 \le 4+2+1 = 7. $$ Thus, $$ |x^3-1| \le 7|x-1|. $$ Step 3. To ensure $$7|x-1|<\epsilon$$, require $$ |x-1| < \frac{\epsilon}{7}. $$ Step 4. Let $$ \delta = \min\left\{1, \frac{\epsilon}{7}\right\}. $$ Then, if $$0 < |x-1| < \delta$$, we have $$ |x^3-1| \le 7|x-1| < 7\cdot \frac{\epsilon}{7} = \epsilon. $$ Thus, by the formal definition of limit, $$\displaystyle \lim_{x \to 1}x^3 = 1$$. --- These proofs complete the epsilon–delta arguments for each limit. **8.1.** As $$x$$ approaches 1, $$2x - 3$$ approaches -1. **8.2.** As $$x$$ approaches -2, $$2x - 3$$ approaches -7. **8.3.** As $$x$$ approaches 1, $$x^3$$ approaches 1.

Une the jrocine (formal defition) of the linit to peove that
8.1. 8.2.

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Calculus Questions

Une the jrocine (formal defition) of the linit to peove that 8.1. 8.2.