Questions: 1. Given that the limit of f(x) as x approaches 3 is 16 and the limit of g(x) as x approaches 3 is -3, find the following limits. (a) The limit of (2f(x) + 10g(x)) as x approaches 3 [2 pts] (b) The limit of (g(x)^2 + 1) / (f(x)^(3/2)) as x approaches 3 [2 pts]

Solution Steps

To solve the given limits, we can use the properties of limits. Specifically, we can use the linearity of limits and the limit of a quotient.

1. For part (a), we use the linearity of limits: \[ \lim _{x \rightarrow 3}(2 f(x) + 10 g(x)) = 2 \lim _{x \rightarrow 3} f(x) + 10 \lim _{x \rightarrow 3} g(x) \] Substitute the given limits: \[ = 2 \cdot 16 + 10 \cdot (-3) \]

2. For part (b), we use the limit of a quotient: \[ \lim _{x \rightarrow 3} \frac{g(x)^{2} + 1}{f(x)^{3 / 2}} = \frac{\lim _{x \rightarrow 3} (g(x)^{2} + 1)}{\lim _{x \rightarrow 3} (f(x)^{3 / 2})} \] Substitute the given limits: \[ = \frac{(-3)^{2} + 1}{16^{3 / 2}} \]

Given: \[ \lim_{x \to 3} f(x) = 16 \quad \text{and} \quad \lim_{x \to 3} g(x) = -3 \]

Using the linearity of limits: \[ \lim_{x \to 3} (2 f(x) + 10 g(x)) = 2 \lim_{x \to 3} f(x) + 10 \lim_{x \to 3} g(x) \]

Substitute the given limits: \[ = 2 \cdot 16 + 10 \cdot (-3) = 32 - 30 = 2 \]

Using the limit of a quotient: \[ \lim_{x \to 3} \frac{g(x)^2 + 1}{f(x)^{3/2}} = \frac{\lim_{x \to 3} (g(x)^2 + 1)}{\lim_{x \to 3} (f(x)^{3/2})} \]

Substitute the given limits: \[ \lim_{x \to 3} (g(x)^2 + 1) = (-3)^2 + 1 = 9 + 1 = 10 \] \[ \lim_{x \to 3} (f(x)^{3/2}) = 16^{3/2} = (16^{1/2})^3 = 4^3 = 64 \]

Thus: \[ \lim_{x \to 3} \frac{g(x)^2 + 1}{f(x)^{3/2}} = \frac{10}{64} = \frac{5}{32} \approx 0.15625 \]

Final Answer