Analyze lim x→∞ f(x) and lim x→−∞ f(x), and then identify any horizontal asymptotes.

f(x)=√x^2+2x+6−3 / x−1

Guida verificata passo dopo passo

Step 1: Identify the dominant terms in the numerator and the denominator as x approaches infinity. For the function \( f(x) = \frac{\sqrt{x^2 + 2x + 6} - 3}{x - 1} \), the dominant term in the numerator is \( \sqrt{x^2} = x \) and in the denominator is \( x \).

Step 2: Simplify the expression by dividing both the numerator and the denominator by the dominant term \( x \). This gives \( \frac{\sqrt{x^2 + 2x + 6}/x - 3/x}{x/x - 1/x} \).

Step 3: Simplify further by recognizing that \( \sqrt{x^2 + 2x + 6}/x = \sqrt{1 + 2/x + 6/x^2} \). As \( x \to \infty \), \( 2/x \to 0 \) and \( 6/x^2 \to 0 \), so \( \sqrt{1 + 2/x + 6/x^2} \to 1 \).

Step 4: Evaluate the limit as \( x \to \infty \). The expression simplifies to \( \frac{1 - 0}{1 - 0} = 1 \). Therefore, \( \lim_{x \to \infty} f(x) = 1 \).

Step 5: Evaluate the limit as \( x \to -\infty \). The dominant term in the numerator becomes \( -x \) because \( \sqrt{x^2} = |x| \) and \( x \) is negative. Simplifying the expression similarly, we find \( \lim_{x \to -\infty} f(x) = -1 \). Thus, the horizontal asymptotes are \( y = 1 \) and \( y = -1 \).

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