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Domain and Range of Trigonometric Ratios

limx→∞sin -1s...

Question

$lim x \to \infty {sin}^{- 1} sin (\frac{x^{2} + 2}{x^{2} + 1})$ is equal to (where [.] and {.} denote greatest integer and fractional part function respectively)

lim x \to \infty [\frac{x^{2} + 2}{x^{2} + 1}]

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lim x \to \infty \frac{x^{2} + 2}{x^{2} + 1}

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lim x \to \infty {\frac{x^{2} + 2}{x^{2} + 1}}

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lim x \to \infty {\frac{x^{2}}{x^{2} + 1}}

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Solution

The correct options are
A $lim x \to \infty [\frac{x^{2} + 2}{x^{2} + 1}]$
B $lim x \to \infty \frac{x^{2} + 2}{x^{2} + 1}$
C $lim x \to \infty {\frac{x^{2}}{x^{2} + 1}}$
$lim x \to \infty {sin}^{- 1} sin (\frac{x^{2} + 2}{x^{2} + 1}) = lim x \to \infty {sin}^{- 1} sin (\frac{1 + 2 / x^{2}}{1 + 1 / x^{2}})$

as $x \to \infty \frac{1}{x^{2}} \to 0$

$= {sin}^{- 1} sin (1) = 1$

A. $lim x \to \infty [\frac{x^{2} + 2}{x^{2} + 1}] = lim x \to \infty [1 + \frac{1}{x^{2} + 1}]$

as $x \to \infty \frac{1}{x^{2} + 1} \to 0$
$= [1] = 1$

B. $lim x \to \infty [\frac{x^{2} + 2}{x^{2} + 1}] = lim x \to \infty \frac{1 + 2 / x^{2}}{1 + 1 / x^{2}}$

$= \frac{1 + 0}{1 + 0} = 1$

C. $lim x \to \infty {\frac{x^{2} + 2}{x^{2} + 1}} = lim x \to \infty (\frac{x^{2} + 2}{x^{2} + 1} - [\frac{x^{2} + 2}{x^{2} + 1}]) ({x} = x - [x])$

From above results

$= 1 - 1 = 0$

D. $lim x \to \infty {\frac{x^{2}}{x^{2} + 1}} = lim x \to \infty (\frac{x^{2}}{x^{2} + 1} - [\frac{x^{2}}{x^{2} + 1}])$

$= {lim}_{x \to \infty} (\frac{1}{1 + 1 / x^{2}} - [1 - \frac{1}{x^{2} + 1}]) (0 < (1 - \frac{1}{x^{2} - 1}) < 1) \Rightarrow [1 - \frac{1}{x^{2} - 1}] = 0$

$= 1 - 0 = 1$

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