How far does √(x² + 6x) end up above x when x → ∞? That is, find the limit of their difference, as an integer.

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How far does $x^{2} + 6 x$ end up above $x$ when $x \to \infty$ ? That is, find the limit of their difference, as an integer.

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Hint

Multiply and divide by the conjugate expression.

Solution

Multiplying by $\frac{x ^{2} + 6 x + x}{x ^{2} + 6 x + x}$ gives $\frac{6 x}{x ^{2} + 6 x + x}$ .
Divide the numerator and the denominator by $x$ to get $\frac{6}{1 + \frac{6}{x} + 1}$ .
As $x \to \infty$ , the root tends to $1$ , so the limit is $\frac{6}{1 + 1} = 3$ .

Answer: $3$