Limits Properties — Question 5

Question 5

Let $f(x) = \sqrt{4x + 5}$ and $g(x) = \dfrac{x^2 + 2x - 3}{x - 1}$, and assume $\lim_{x \to 1} f(x) = 3$.
(a) Find $\displaystyle \lim_{x \to 1} g(x)$.
(b) Use the limit properties to find $\displaystyle \lim_{x \to 1} [f(x) \cdot g(x)]$.

Question 5 - Solution

We are given:

$$f(x) = \sqrt{4x + 5}, \quad g(x) = \dfrac{x^2 + 2x - 3}{x - 1}, \quad \lim_{x \to 1} f(x) = 3$$

(a) Find $\lim_{x \to 1} g(x)$:

First, factor the numerator of $g(x)$:

$$x^2 + 2x - 3 = (x + 3)(x - 1)$$ $$g(x) = \frac{(x + 3)(x - 1)}{x - 1}$$

For $x \neq 1$, cancel common factors: $$g(x) = x + 3$$

So, $$\lim_{x \to 1} g(x) = \lim_{x \to 1} (x + 3) = 4$$

Answer: $\boxed{4}$

(b) Find $\lim_{x \to 1} [f(x) \cdot g(x)]$:

Use the limit product law: $$\lim_{x \to 1} [f(x) \cdot g(x)] = \lim_{x \to 1} f(x) \cdot \lim_{x \to 1} g(x) = 3 \cdot 4 = 12$$

Answer: $\boxed{12}$