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Let's go through the problems one by one, starting from 12 to 17.

### 12. \( \lim_{h \to 0} \frac{3(1 + h)^2 - 7(1 + h) + 1 + 3}{h} \)

This is a difference quotient, likely the derivative of a function evaluated at a point. To find the function \( f(x) \), we can expand the numerator:

\[
3(1 + h)^2 = 3(1 + 2h + h^2) = 3 + 6h + 3h^2
\]
\[
-7(1 + h) = -7 - 7h
\]
\[
1 + 3 = 4
\]

Now combine terms:

\[
3 + 6h + 3h^2 - 7 - 7h + 4 = (0) + (-h) + 3h^2
\]

So the expression becomes:

\[
\frac{-h + 3h^2}{h} = -1 + 3h
\]

Taking the limit as \( h \to 0 \):

\[
\lim_{h \to 0} (-1 + 3h) = -1
\]

The function \( f(x) = 3x^2 - 7x + 1 \), and the value of \( c \) where \( f'(c) = -1 \) is \( c = 1 \).

---

### 13. \( \lim_{h \to 0} \frac{\log(2 - 4(h - 5)) - \log(22)}{h} \)

We first simplify the expression inside the logarithm:

\[
2 - 4(h - 5) = 2 - 4h + 20 = 22 - 4h
\]

So the limit becomes:

\[
\lim_{h \to 0} \frac{\log(22 - 4h) - \log(22)}{h}
\]

Using the logarithmic property \( \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \):

\[
\lim_{h \to 0} \frac{\log\left(\frac{22 - 4h}{22}\right)}{h}
= \lim_{h \to 0} \frac{\log\left(1 - \frac{4h}{22}\right)}{h}
\]

For small \( h \), use the approximation \( \log(1 + x) \approx x \) for \( x \) near 0:

\[
\lim_{h \to 0} \frac{-\frac{4h}{22}}{h} = -\frac{4}{22} = -\frac{2}{11}
\]

Thus, \( f'(c) = -\frac{2}{11} \), and the function \( f(x) = \log(22 - 4x) \), evaluated at \( c = 5 \).

---

### 14. \( \lim_{x \to -2} \frac{3x - 9x^2 + 42}{x + 2} \)

Substitute \( x = -2 \):

\[
3(-2) - 9(-2)^2 + 42 = -6 - 36 + 42 = 0
\]

Both the numerator and denominator equal 0, indicating the possibility of factoring or using L'Hopital's Rule.

First, factor the numerator:

\[
3x - 9x^2 + 42 = -9(x^2 - \frac{1}{3}x - \frac{14}{9})
\]

We can attempt polynomial division or L'Hopital's rule directly, which would give the result as \( 0 \), implying simplification error.

---

### 15. \( \lim_{x \to 5} \left( \frac{1}{\sqrt{3x}} - \frac{1}{\sqrt{15}} \right) / (x - 5) \)

This can be solved using L'Hopital's Rule, or by rationalizing the numerator. Let's use L'Hopital's Rule:

First, take the derivative of the numerator and denominator:

- Derivative of the numerator: \( \frac{d}{dx} \left( \frac{1}{\sqrt{3x}} - \frac{1}{\sqrt{15}} \right) = \frac{-3}{2(3x)^{3/2}} \)
- Derivative of the denominator: \( 1 \)

Substitute \( x = 5 \) into the derivative:

\[
\lim_{x \to 5} \frac{-3}{2(3(5))^{3/2}} = -\frac{3}{30\sqrt{5}}
\]

---

### 16. \( \lim_{h \to 0} \frac{e^{6(3 + h)} + 1 - e^{19}}{h} \)

Simplify the exponentials:

\[
e^{6(3 + h)} = e^{18 + 6h} = e^{18} \cdot e^{6h}
\]

Now the limit becomes:

\[
\lim_{h \to 0} \frac{e^{18} \cdot e^{6h} + 1 - e^{19}}{h}
\]

Since \( e^{19} = e^{18} \cdot e \), we can apply L'Hopital’s Rule or approximate for small \( h \). Using L'Hopital's rule, differentiate the numerator and denominator.

---

### 17. \( \lim_{x \to \frac{\pi}{2}} \frac{6x^2 \sin \left( x - \frac{3}{2}\pi^2 \right)}{x - \frac{\pi}{2}} \)

This is another case where L'Hopital's Rule or Taylor expansion can be used to evaluate the limit. Differentiating the numerator and denominator:



Identify the original function f(x), and what value of c to evaluate f'(c) for the given limits.

This page provides a step-by-step solution for identifying the original function f(x) and the derivative f'(c) by evaluating limit problems. Concepts include limits, derivatives, logarithmic properties, and L'Hopital's Rule.

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