Math · Calculus I · Worked example

Find a limit at infinity with the conjugate

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

Recognize the form

Both terms grow without bound, so the difference has the indeterminate form ∞ − ∞. Its limit could be anything, so more work is needed.

Multiply and divide by √(x² + 4x) + x. The numerator becomes (x² + 4x) − x² = 4x, so for x > 0 the expression equals 4x/(√(x² + 4x) + x).

For x > 0, √(x² + 4x) = x√(1 + 4/x), so the quotient is 4/(√(1 + 4/x) + 1), which tends to 4/(1 + 1) = 2.

The values creep up toward 2. Numbers suggest a limit; the algebra above proves it.

\begin{array}{l} \sqrt{100^{2} + 400} - 100 \approx 1.98 \\ \sqrt{10000^{2} + 40000} - 10000 \approx 2.00 \end{array}

The limit is 2, even though both terms grow without bound.

Find the limit of √(x² + 6x) − x as x → ∞.

Show the answer and explanation

The conjugate gives 6x/(√(x² + 6x) + x), which equals 6/(√(1 + 6/x) + 1) for x > 0 and tends to 6/2 = 3.

\sqrt{10000^{2} + 60000} - 10000 \approx 3.00

Graph plots √(x² + 4x) − x with the line y = 2; zoom out to watch the curve level off.

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

Make it concrete

Open the example inputs, change a value and keep a useful result on your board.

Your existing work stays on this device. Examples open as editable copies.