Horizontal and Vertical Asymptotes


Use limits to determine the equations for all horizontal asymptotes.
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√x² + 4
c. g(x) = -----------
x

Verified step by step guidance

1. Identify the function: \( g(x) = \frac{\sqrt{x^2 + 4}}{x} \). We need to find the horizontal asymptotes by evaluating the limits as \( x \to \infty \) and \( x \to -\infty \).
2. Consider the limit as \( x \to \infty \). Simplify the expression \( \sqrt{x^2 + 4} \) by factoring out \( x^2 \) from under the square root: \( \sqrt{x^2 + 4} = \sqrt{x^2(1 + \frac{4}{x^2})} = x\sqrt{1 + \frac{4}{x^2}} \).
3. Substitute the simplified form back into the function: \( g(x) = \frac{x\sqrt{1 + \frac{4}{x^2}}}{x} = \sqrt{1 + \frac{4}{x^2}} \).
4. Evaluate the limit as \( x \to \infty \): \( \lim_{x \to \infty} \sqrt{1 + \frac{4}{x^2}} = \sqrt{1 + 0} = 1 \). Thus, there is a horizontal asymptote at \( y = 1 \) as \( x \to \infty \).
5. Similarly, evaluate the limit as \( x \to -\infty \). The simplification process is the same, and the limit \( \lim_{x \to -\infty} \sqrt{1 + \frac{4}{x^2}} = 1 \) also holds. Therefore, there is a horizontal asymptote at \( y = 1 \) as \( x \to -\infty \).