The value of lim x - 0-miny2-6 y-17 sin x-x- denotes the greatest integer function

The correct option is B 7min(y^2+6y+17)= min((y+3)^2+8)=8 L=lim_x → 0[min(y^2+6y+17)sin xx] =lim_x → 0[8sin xx] =[a value slightly less than 8] nbsp; [∵ |s ...

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Standard XII

Mathematics

Real Valued Functions

The value of ...

Question

The value of $lim x \to 0 [min (y^{2} + 6 y + 17) \frac{sin x}{x}]$ is
$([.]$ denotes the greatest integer function $)$

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Solution

The correct option is C $7$
$min (y^{2} + 6 y + 17) = min ((y + 3)^{2} + 8) = 8$

$L = lim x \to 0 [min (y^{2} + 6 y + 17) \frac{sin x}{x}]$

$= lim x \to 0 [\frac{8 sin x}{x}]$
$= [a value slightly less than 8]$ $[∵ | sin x | < | x |$ when $x \neq 0]$
$∴ L = 7$

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The value of limx→0[min(y2+6y+17)sinxx] is ( [.] denotes the greatest integer function)

The correct option is C 7min(y2+6y+17)=min((y+3)2+8)=8 L=limx→0[min(y2+6y+17)sinxx] =limx→0[8sinxx] =[a value slightly less than 8] [∵|sinx|<|x| when x≠0] ∴L=7

The value of $lim x \to 0 [min (y^{2} + 6 y + 17) \frac{sin x}{x}]$ is
$([.]$ denotes the greatest integer function $)$

The correct option is C $7$
$min (y^{2} + 6 y + 17) = min ((y + 3)^{2} + 8) = 8$

$L = lim x \to 0 [min (y^{2} + 6 y + 17) \frac{sin x}{x}]$

$= lim x \to 0 [\frac{8 sin x}{x}]$
$= [a value slightly less than 8]$ $[∵ | sin x | < | x |$ when $x \neq 0]$
$∴ L = 7$