8. Use the precise (formal defition) of the limit to prove that 8.1. $$ \lim _{x \rightarrow 3}(2 x-3)=3 $$ 8.2. $$ \lim _{x \rightarrow-3}(2 x-3)=-9 \quad $$ 8.3. $$ \lim _{x \rightarrow 2} x^{2}=4 $$

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**8.1. Prove that $$ \lim_{x \rightarrow 3}(2x-3)=3 $$.** We must show that for every $$ \epsilon > 0 $$ there exists a $$ \delta > 0 $$ such that if $$ |x-3| < \delta \quad \text{then} \quad |(2x-3) - 3| < \epsilon. $$ 1. Compute the difference: $$ |(2x-3) - 3| = |2x - 6| = 2|x-3|. $$ 2. To ensure that $$ 2|x-3| < \epsilon $$, it is sufficient to require $$ |x-3| < \frac{\epsilon}{2}. $$ 3. Hence, choose $$ \delta = \frac{\epsilon}{2}. $$ 4. Then, for any $$ x $$ satisfying $$ |x-3| < \delta $$, we have $$ |(2x-3) - 3| = 2|x-3| < 2\left(\frac{\epsilon}{2}\right) = \epsilon, $$ which completes the proof. --- **8.2. Prove that $$ \lim_{x \rightarrow -3}(2x-3)=-9 $$.** We want to show: for every $$ \epsilon > 0 $$ there exists a $$ \delta > 0 $$ such that if $$ |x - (-3)| < \delta \quad \text{then} \quad |(2x-3) - (-9)| < \epsilon. $$ 1. Write the difference: $$ |(2x-3) - (-9)| = |2x - 3 + 9| = |2x + 6| = 2|x+3|. $$ 2. To ensure $$ 2|x+3| < \epsilon $$ we need $$ |x+3| < \frac{\epsilon}{2}. $$ 3. Therefore, set $$ \delta = \frac{\epsilon}{2}. $$ 4. Then, for any $$ x $$ such that $$ |x+3| < \delta $$, $$ |(2x-3) - (-9)| = 2|x+3| < 2\left(\frac{\epsilon}{2}\right) = \epsilon. $$ This completes the proof. --- **8.3. Prove that $$ \lim_{x \rightarrow 2} x^2=4 $$.** We must show that for every $$ \epsilon > 0 $$ there exists a $$ \delta > 0 $$ such that if $$ |x-2| < \delta \quad \text{then} \quad |x^2-4| < \epsilon. $$ 1. Factor the difference: $$ |x^2 - 4| = |(x-2)(x+2)| = |x-2| \cdot |x+2|. $$ 2. To control the factor $$ |x+2| $$, choose a preliminary bound. Assume $$ \delta \leq 1. $$ Then if $$ |x-2| < 1 $$, it follows that $$ 1 < x < 3 \quad \text{and hence} \quad 3 < x+2 < 5. $$ Thus, we have $$ |x+2| < 5. $$ 3. Now, using this bound, we have: $$ |x^2-4| = |x-2| \cdot |x+2| < 5|x-2|. $$ 4. To ensure $$ |x^2-4| < \epsilon $$, it is enough to require: $$ 5|x-2| < \epsilon \quad \Rightarrow \quad |x-2| < \frac{\epsilon}{5}. $$ 5. Now, choose $$ \delta = \min\left\{1, \frac{\epsilon}{5}\right\}. $$ 6. Then, if $$ |x-2| < \delta $$, $$ |x^2-4| < 5|x-2| < 5\left(\frac{\epsilon}{5}\right) = \epsilon. $$ This completes the proof. **8.1. $$ \lim_{x \rightarrow 3}(2x-3)=3 $$** For every $$ \epsilon > 0 $$, choose $$ \delta = \frac{\epsilon}{2} $$. If $$ |x - 3| < \delta $$, then $$ |(2x-3) - 3| < \epsilon $$. **8.2. $$ \lim_{x \rightarrow -3}(2x-3)=-9 $$** For every $$ \epsilon > 0 $$, choose $$ \delta = \frac{\epsilon}{2} $$. If $$ |x + 3| < \delta $$, then $$ |(2x-3) - (-9)| < \epsilon $$. **8.3. $$ \lim_{x \rightarrow 2} x^2=4 $$** For every $$ \epsilon > 0 $$, choose $$ \delta = \min\left\{1, \frac{\epsilon}{5}\right\} $$. If $$ |x - 2| < \delta $$, then $$ |x^2 - 4| < \epsilon $$.

Use the precise (formal defition) of the limit to prove that
8.1.
8.2. 8.3.

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Use the precise (formal defition) of the limit to prove that 8.1. 8.2. 8.3.