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Problem 12:

We are asked to find $\lim_{x \to 5} f(g(x))$, given the following information:

• $\lim_{y \to 7} f(y) = 5$,
• $f(7) = 5$,
• $f(5)$ is undefined,
• $f$ is continuous,
• $\lim_{x \to 5} g(x) = 7$.

To solve this, we use the fact that $g(x) \to 7$ as $x \to 5$. Since $f$ is continuous and $\lim_{y \to 7} f(y) = 5$, we can substitute the limit of $g(x)$ into $f(g(x))$: $\lim_{x \to 5} f(g(x)) = f\left(\lim_{x \to 5} g(x)\right) = f(7) = 5.$ Thus, $\lim_{x \to 5} f(g(x)) = 5$.

Exercise 22:

The function given is: $f(x) = \frac{x^2 - 9}{x^3 + 2x^2 - 9x - 18} = \frac{(x-3)(x+3)}{(x-3)(x+2)(x+3)}.$ Here, we need to determine the points of discontinuity.

1. Factor the function: As we can see, the numerator and denominator have common factors of $(x-3)$ and $(x+3)$. This simplifies the function to: $f(x) = \frac{1}{x+2}, \quad x \neq 3, x \neq -3.$ So, the points of interest are $x = 3$ and $x = -3$.

2. Check discontinuities:

   • At $x = 3$, there is a removable discontinuity because the factor $(x - 3)$ cancels out.
   • At $x = -3$, the function has a vertical asymptote because the factor $(x + 3)$ does not cancel in the denominator.

Thus, $x = 3$ is a removable discontinuity, and $x = -3$ is a vertical asymptote (non-removable discontinuity).

Exercise 23:

The piecewise function $g(x)$ is given as: