If $lim x \to 2^{-} \frac{\sqrt{4 - x^{2}} \sqrt{cos \frac{π x}{4}}}{sin (x - 2)} = k$ and $[.]$ denotes the greatest integer function, then the value of $[k + 5]$ is

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Solution

$lim x \to 2^{-} \frac{\sqrt{4 - x^{2}} \sqrt{cos \frac{π x}{4}}}{sin (x - 2)}$
$= lim x \to 2^{-} \frac{\sqrt{(2 + x) (2 - x)} \sqrt{cos \frac{π x}{4}}}{sin (x - 2)}$
Put $x = 2 - h$
$k = lim h \to 0 \frac{\sqrt{h (4 - h)} \sqrt{cos (\frac{π}{2} - \frac{π h}{4})}}{- sin h}$
$= lim h \to 0 \frac{\sqrt{h (4 - h)} \sqrt{sin \frac{π h}{4}}}{- sin h}$
$= lim h \to 0 \frac{\sqrt{h (4 - h)} \sqrt{\frac{sin (π h / 4)}{(π h / 4)} \times (π h / 4)}}{- \frac{sin h}{h} \times h}$
$= lim h \to 0 \frac{\sqrt{(4 - h)} \sqrt{\frac{π}{4}}}{- 1} = - \sqrt{π}$
$[- \sqrt{π}] = - 2$
$∴ [k + 5] = [k] + 5 = - 2 + 5 = 3$

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