Horizontal asymptotes of a rational-radical function

Consider the function $f(x) = \dfrac{\sqrt{9x^2+1}}{x+2}$ .

Find $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty} f(x)$ , and use these results to state the equation(s) of all horizontal asymptote(s) of $f(x) = \dfrac{\sqrt{9x^2+1}}{x+2}$ .

Hint

Divide numerator and denominator by $x$ (the highest power in the denominator). Remember that when pulling $x$ out of $\sqrt{9x^2+1}$ , you must write $x = \sqrt{x^2}$ , and be careful with the sign of $x$ as $x \to -\infty$ .

Step-by-step solution

Divide numerator and denominator by $x$ . For $x>0$ , write $\sqrt{9x^2+1} = x\sqrt{9 + 1/x^2}$ since $x = \sqrt{x^2}$ is positive. Then $f(x) = \frac{x\sqrt{9+1/x^2}}{x(1+2/x)} = \frac{\sqrt{9+1/x^2}}{1+2/x}.$ As $x \to \infty$ , $1/x^2 \to 0$ and $2/x \to 0$ , so the expression tends to $\sqrt{9}/1 = 3$ .

For $x<0$ , $x = -\sqrt{x^2}$ , so $\sqrt{9x^2+1} = |x|\sqrt{9+1/x^2} = -x\sqrt{9+1/x^2}$ . Dividing numerator and denominator by $x$ (a negative number) flips the sign: $f(x) = \frac{-x\sqrt{9+1/x^2}}{x(1+2/x)} = \frac{-\sqrt{9+1/x^2}}{1+2/x}.$ As $x \to -\infty$ , this tends to $-\sqrt{9}/1 = -3$ . Thus the two limits differ, giving two horizontal asymptotes: $y=3$ (as $x\to\infty$ ) and $y=-3$ (as $x\to-\infty$ ).

Horizontal asymptotes of a rational-radical function — solution graph

Horizontal asymptotes of a rational-radical function

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