Math · Calculus I · Worked example

Find horizontal asymptotes at both ends

Find the limits of f(x) = 2x/√(x² + 1) as x → ∞ and as x → −∞.

Factor x² out of the root

√(x² + 1) = √(x²)·√(1 + 1/x²) = |x|√(1 + 1/x²). The absolute value is the whole point.

For x > 0, |x| = x, so f(x) = 2/√(1 + 1/x²), which tends to 2/√1 = 2.

For x < 0, |x| = −x, so f(x) = −2/√(1 + 1/x²), which tends to −2.

Values at x = ±100 agree with the limits.

\begin{array}{l} f (x) = \frac{2 x}{\sqrt{x^{2} + 1}} \\ f (100) \approx 2.00 \\ f (- 100) \approx - 2.00 \end{array}

f(x) → 2 as x → ∞ and f(x) → −2 as x → −∞: the lines y = 2 and y = −2 are both horizontal asymptotes.

Find the limit of 3x/√(4x² + 5) as x → −∞.

Show the answer and explanation

For x < 0, √(4x² + 5) = −x√(4 + 5/x²), so the quotient equals −3/√(4 + 5/x²), which tends to −3/2.

\begin{array}{l} g (x) = \frac{3 x}{\sqrt{4 x^{2} + 5}} \\ g (- 1000) \approx - 1.50 \end{array}

Graph plots f with the lines y = 2 and y = −2.

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