$$  $$$$ \lim_{h\overrightarrow{}0} $$$$ \frac{f\left(3+h\right)-f\left(3\right)}{h} $$does the lim as x approaches 5 exist, is f continouous on the closed interval -1,7 and for some c, 0<c<7, f(c)=-2

Let's break down the problem step by step, starting with each part separately.

### 1. Evaluate the limit expression:
The limit expression given is:

\[
\lim_{h \to 0} \frac{f(3+h) - f(3)}{h}
\]

This limit represents the derivative of the function $ f(x) $ at $ x = 3 $. Specifically, it is:

\[
f'(3) = \lim_{h \to 0} \frac{f(3+h) - f(3)}{h}
\]

To determine whether this limit exists, we need more information about the function $ f(x) $. If $ f(x) $ is differentiable at $ x = 3 $, then this limit exists and equals $ f'(3) $.

### 2. Does the limit as $ x $ approaches 5 exist?

The question asks whether $ \lim_{x \to 5} f(x) $ exists. For this limit to exist, the function $ f(x) $ must approach a specific value as $ x $ gets arbitrarily close to 5 from both sides (left and right). Again, this depends on the behavior of $ f(x) $ near $ x = 5 $.

### 3. Is $ f(x) $ continuous on the closed interval $[-1, 7]$?

A function $ f(x) $ is continuous on a closed interval $[-1, 7]$ if:

- $ f(x) $ is continuous at every point in the open interval $(-1, 7)$.
- $ f(x) $ is continuous from the right at $ x = -1 $.
- $ f(x) $ is continuous from the left at $ x = 7 $.

To determine continuity, we need to check if the limits $ \lim_{x \to c} f(x) $ exist and equal $ f(c) $ for all $ c $ in $[-1, 7]$.

### 4. Is there a $ c $ in $ (0, 7) $ such that $ f(c) = -2 $?

This part of the problem is related to the Intermediate Value Theorem (IVT). The IVT states that if $ f(x) $ is continuous on a closed interval $[a, b]$, and $ N $ is any number between $ f(a) $ and $ f(b) $, then there exists at least one number $ c $ in the open interval $ (a, b) $ such that $ f(c) = N $.

To answer whether there exists some $ c $ in $ (0, 7) $ such that $ f(c) = -2 $, we need to know:

- If $ f(x) $ is continuous on $[0, 7]$.
- The values $ f(0) $ and $ f(7) $.

If $ f(x) $ is continuous on $[0, 7]$ and $ -2 $ is between $ f(0) $ and $ f(7) $, then by the IVT, there exists some $ c $ in $ (0, 7) $ such that $ f(c) = -2 $.

### Summary:
To fully answer the questions posed, we need more specific information about the function $ f(x) $. If provided, we can determine:

1. Whether $ f'(3) $ exists (the derivative at $ x = 3 $).
2. Whether the limit $ \lim_{x \to 5} f(x) $ exists.
3. Whether $ f(x) $ is continuous on the interval $[-1, 7]$.
4. Whether there is a $ c $ such that $ f(c) = -2 $ in the interval $ (0, 7) $.

Would you like to provide more details on $ f(x) $ or explore these concepts further?

Here are 5 related questions:
1. What are the conditions for a function to be differentiable at a point?
2. How do you determine if a limit exists at a particular point?
3. What is the significance of the Intermediate Value Theorem in calculus?
4. How can you check the continuity of a function on a closed interval?
5. What is the relationship between differentiability and continuity?

**Tip:** Always check the continuity of a function before applying the Intermediate Value Theorem.

Explore the analysis of limits, continuity, and derivatives in function analysis. Understand the existence of limits as x approaches 5, continuity on the closed interval [-1, 7], and the application of the Intermediate Value Theorem. Get insights into solving complex mathematical problems involving functions and their behavior.

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