We are asked to prove, using the precise definition of a limit, that $\lim_{x \to 10} (3 - \frac{4}{5}x) = -5$ and $\lim_{x \to -1^-} \frac{5}{(x+1)^3} = -\infty$.

1. Problem Description

We are asked to prove, using the precise definition of a limit, that $\lim_{x \to 10} (3 - \frac{4}{5}x) = -5$ and $\lim_{x \to -1^-} \frac{5}{(x+1)^3} = -\infty$.

2. Solution Steps

2.

2. Proving $\lim_{x \to 10} (3 - \frac{4}{5}x) = -5$

We need to show that for every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - 10| < \delta$, then $|(3 - \frac{4}{5}x) - (-5)| < \epsilon$.

We have:

$|(3 - \frac{4}{5}x) - (-5)| = |3 - \frac{4}{5}x + 5| = |8 - \frac{4}{5}x| = |-\frac{4}{5}(x - 10)| = \frac{4}{5}|x - 10|$.

We want $\frac{4}{5}|x - 10| < \epsilon$. This means $|x - 10| < \frac{5}{4}\epsilon$.

Let $\delta = \frac{5}{4}\epsilon$. Then, if $0 < |x - 10| < \delta = \frac{5}{4}\epsilon$, we have

$|(3 - \frac{4}{5}x) - (-5)| = \frac{4}{5}|x - 10| < \frac{4}{5}(\frac{5}{4}\epsilon) = \epsilon$.

Thus, for any $\epsilon > 0$, we can choose $\delta = \frac{5}{4}\epsilon$, and the precise definition of the limit is satisfied.

2.

3. Proving $\lim_{x \to -1^-} \frac{5}{(x+1)^3} = -\infty$

We need to show that for every $M > 0$, there exists a $\delta > 0$ such that if $-1 - \delta < x < -1$, then $\frac{5}{(x+1)^3} < -M$.

Since $x < -1$, we have $x+1 < 0$, and therefore $(x+1)^3 < 0$. Also, we consider $x$ approaching $-1$ from the left, i.e., $x \to -1^-$. We want $\frac{5}{(x+1)^3} < -M$. This is equivalent to $(x+1)^3 > -\frac{5}{M}$.

Taking the cube root of both sides, we have $x+1 > -\sqrt[3]{\frac{5}{M}}$.

This means $x > -1 - \sqrt[3]{\frac{5}{M}}$.

We are given that $-1 - \delta < x < -1$. Therefore, we can choose $\delta = \sqrt[3]{\frac{5}{M}}$. Then, $-1 - \delta < x < -1$ implies $-1 - \sqrt[3]{\frac{5}{M}} < x < -1$.

Thus, $x+1 > -\sqrt[3]{\frac{5}{M}}$, so $(x+1)^3 > -\frac{5}{M}$, and therefore $\frac{5}{(x+1)^3} < -M$.

Therefore, for any $M > 0$, we can choose $\delta = \sqrt[3]{\frac{5}{M}}$, and the precise definition of the limit at negative infinity is satisfied.

3. Final Answer

2.

2. For every $\epsilon > 0$, choose $\delta = \frac{5}{4}\epsilon$.

2.

3. For every $M > 0$, choose $\delta = \sqrt[3]{\frac{5}{M}}$.