Limit lim x → 0 sin ⁡ [ x ] [ x ]

The domain of the function excludes [0,1), so that lim x → 0 f ( x ) = lim x → 0 − f ( x ) = − sin ⁡ ( − 1 ) .

For − 1 ≤ x &lt; 0, you have [ x ] = − 1, so for x in that interval: sin ⁡ [ x ] [ x ] = sin ⁡ ( − 1 ) − 1 = sin ⁡ 1Which makes the left-handed limit sin ⁡ 1For 0 ≤ x &lt; 1, you have [x]=0 so sin ⁡ x x is not defined in neighborhoods on the right of x=0.

The domain of the function excludes [0,1), so that lim x → 0 f ( x ) = lim x → 0 − f ( x ) = − sin ⁡ ( − 1 ) .

For − 1 ≤ x &lt; 0, you have [ x ] = − 1, so for x in that interval: sin ⁡ [ x ] [ x ] = sin ⁡ ( − 1 ) − 1 = sin ⁡ 1Which makes the left-handed limit sin ⁡ 1For 0 ≤ x &lt; 1, you have [x]=0 so sin ⁡ x x is not defined in neighborhoods on the right of x=0.

Find out the answer to "Limit lim x → 0 sin ⁡ [..."

bandikizaui

Answered question

2022-07-04

Limit

lim_{x \to 0} \frac{\sin [x]}{[x]}

Answer & Explanation

Leslie Rollins

Beginner2022-07-05Added 25 answers

The domain of the function excludes [0,1), so that

lim_{x \to 0} f (x) = lim_{x \to 0^{-}} f (x) = - \sin (- 1) .

kramberol

Beginner2022-07-06Added 5 answers

For

- 1 \leq x < 0

, you have

[x] = - 1

, so for x in that interval:

\frac{\sin [x]}{[x]} = \frac{\sin (- 1)}{- 1} = \sin 1

Which makes the left-handed limit

\sin 1

For

0 \leq x < 1

, you have [x]=0 so

\frac{\sin x}{x}

is not defined in neighborhoods on the right of x=0.

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