Use the inequality x - 1 < ⌊ x ⌋ ≤ x together with the Squeeze Theorem to evaluate limx → ∞ ⌊ x ⌋/x. Show the bounding inequality you use for dfrac⌊ x ⌋x, and state the value of the limit.

Starting from x - 1 0 (which preserves the inequality direction) gives x-1/x 0. Now take the limit as x → ∞ of the outer expressions: limx→∞left(1 - 1/xright) = 1 and limx→∞ 1 = 1. Since ⌊ x ⌋/x is squeezed between two expressions that both approach 1$, the Squeeze Theorem gives limx → ∞ ⌊ x ⌋/x = 1.

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Squeeze theorem applied to floor(x)/x as x -> infinity

Recall the floor function $\lfloor x \rfloor$ (the greatest integer less than or equal to $x$ ) satisfies the inequality $x - 1 < \lfloor x \rfloor \le x$ for all real $x$ .

Use the inequality $x - 1 < \lfloor x \rfloor \le x$ together with the Squeeze Theorem to evaluate $\lim_{x \to \infty} \frac{\lfloor x \rfloor}{x}.$ Show the bounding inequality you use for $\dfrac{\lfloor x \rfloor}{x}$ , and state the value of the limit.

Hint

Divide every part of the inequality $x-1 < \lfloor x \rfloor \le x$ by $x$ (valid for $x>0$ ), then take the limit of the resulting lower and upper bounds as $x \to \infty$ .

Step-by-step solution

Starting from $x - 1 < \lfloor x \rfloor \le x$ and dividing through by $x > 0$ (which preserves the inequality direction) gives $\frac{x-1}{x} < \frac{\lfloor x \rfloor}{x} \le 1.$ This rewrites the lower bound as $1 - \frac{1}{x}$ , so we have $1 - \frac{1}{x} < \frac{\lfloor x \rfloor}{x} \le 1$ for all $x > 0$ .

Now take the limit as $x \to \infty$ of the outer expressions: $\lim_{x\to\infty}\left(1 - \frac{1}{x}\right) = 1$ and $\lim_{x\to\infty} 1 = 1$ . Since $\frac{\lfloor x \rfloor}{x}$ is squeezed between two expressions that both approach $1$ , the Squeeze Theorem gives $\lim_{x \to \infty} \frac{\lfloor x \rfloor}{x} = 1.$

Squeeze theorem applied to floor(x)/x as x -> infinity

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