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Limit at infinity of a rational expression involving a square root

Consider the function g(x)=x2+1x.g(x) = \frac{\sqrt{x^2+1}}{x}. Evaluate lim⁡x→∞g(x).\lim_{x \to \infty} g(x). Note that this limit has the indeterminate form ∞/∞\infty/\infty, but rather than applying L'Hopital's Rule directly, it is easier to divide the numerator and denominator by an appropriate power of xx. Determine the value of the limit.

Hint

Since x→∞x \to \infty, you may assume x>0x > 0, so x=x2x = \sqrt{x^2}. Divide numerator and denominator by xx, writing the numerator as x2+1/x=(x2+1)/x2\sqrt{x^2+1}/x = \sqrt{(x^2+1)/x^2}, then take the limit as x→∞x \to \infty.

Step-by-step solution

As x→∞x \to \infty, both x2+1→∞\sqrt{x^2+1} \to \infty and x→∞x \to \infty, so this is an ∞/∞\infty/\infty indeterminate form. Since x→∞x \to \infty, we know x>0x > 0, so we can write x=x2x = \sqrt{x^2} and bring it inside the square root: g(x)=x2+1x=x2+1x2=1+1x2.g(x) = \frac{\sqrt{x^2+1}}{x} = \sqrt{\frac{x^2+1}{x^2}} = \sqrt{1 + \frac{1}{x^2}}.

Now take the limit: as x→∞x \to \infty, 1x2→0\frac{1}{x^2} \to 0, so lim⁡x→∞1+1x2=1+0=1.\lim_{x\to\infty} \sqrt{1+\frac{1}{x^2}} = \sqrt{1+0} = 1. Therefore lim⁡x→∞g(x)=1\lim_{x\to\infty} g(x) = 1, meaning y=1y=1 is a horizontal asymptote of gg as x→∞x\to\infty.

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