I have posted the functions that are needed in a separate picture. Thank you so much Consider the functions $ f(x) = \frac{3x^2 + 16x - 12}{x + 6} $, \[ g(x) = \begin{cases} 3x - 2 & \text{when } x \neq -6 \\ -10 & \text{when } x = -6 \end{cases} \]and $ h(x) = 3x - 2 $. **Problem Statement:**Consider the functions $ f, g, $ and $ h $ from Exercise 2.A. Provide an $\epsilon-\delta$ proof to support the fact that $\lim_{{x \to 6}} h(x)$ is equal to $-20$.B. Consider the statement: "For a positive real number $\delta$, on any interval of the form $(-6 - \delta, -6) \cup (-6, -6 + \delta)$, the function $ f(x) $ is identical to $3x - 2$." Proper use of this statement allows an $\epsilon-\delta$ proof of $\lim_{{x \to -6}} f(x) = -20$ to, in effect, only be one sentence different from that of your proof in part A. Explain. *(Note: you do not need to re-write your result from part A. You just need to explain why this statement in quotes applies to an $\epsilon-\delta$ proof of $\lim_{{x \to -6}} f(x) = -20$ and why it effectively simplifies an $\epsilon-\delta$ argument to that which you wrote in part A.)*C. Similar to part B, an $\epsilon-\delta$ proof of $\lim_{{x \to -6}} g(x) = -20$ can differ from your proof in part A by effectively one sentence. Explain.

Here, in the question consider the functions f(x) = 3x+16x-12x+6, g(x) = {-10 when x =-6 3x-2 when…

Answered: e functions f, g, and h from Exercise…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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I have posted the functions that are needed in a separate picture. Thank you so much

Consider the functions $ f(x) = \frac{3x^2 + 16x - 12}{x + 6} $,

\[ g(x) =
\begin{cases}
3x - 2 & \text{when } x \neq -6 \\
-10 & \text{when } x = -6
\end{cases} \]

and $ h(x) = 3x - 2 $.

**Problem Statement:**

Consider the functions $ f, g, $ and $ h $ from Exercise 2.

A. Provide an $\epsilon-\delta$ proof to support the fact that $\lim_{{x \to 6}} h(x)$ is equal to $-20$.

B. Consider the statement: "For a positive real number $\delta$, on any interval of the form $(-6 - \delta, -6) \cup (-6, -6 + \delta)$, the function $ f(x) $ is identical to $3x - 2$." Proper use of this statement allows an $\epsilon-\delta$ proof of $\lim_{{x \to -6}} f(x) = -20$ to, in effect, only be one sentence different from that of your proof in part A. Explain. *(Note: you do not need to re-write your result from part A. You just need to explain why this statement in quotes applies to an $\epsilon-\delta$ proof of $\lim_{{x \to -6}} f(x) = -20$ and why it effectively simplifies an $\epsilon-\delta$ argument to that which you wrote in part A.)*

C. Similar to part B, an $\epsilon-\delta$ proof of $\lim_{{x \to -6}} g(x) = -20$ can differ from your proof in part A by effectively one sentence. Explain.

Expert Solution

Step 1

Here, in the question consider the functions $f (x) = \frac{3 x + 16 x - 12}{x + 6}, g (x) = {_{- 10_{w h e n x = - 6}}^{3 x - 2 w h e n x = 6}$ and $h (x) = 3 x - 2$

We have to prove for parts A, B and C.

We have to prove for the point the epsilon-delta definition of limit says that the limit of $f (x)$ at $x = c$ is $L$ if for any $ε > 0$ there's a $δ > 0$ such that if the distance of $x$ from $c$ is less than $δ$ then distance of $f (x)$ from $L$ is less than $ε$ .