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Question Number 111960 by john santu last updated on 05/Sep/20

(1) \lim_{x \to \frac{π}{4}} \sin (\frac{π}{4} - x) . \tan (x + \frac{π}{4}) ?

(2) \lim_{x \to \frac{π}{2}} \frac{π (π - 2 x) \tan (x - \frac{π}{2})}{2 (x - π) \cos^{2} x} ?

(3) \lim_{x \to 0} \frac{3 x (\cos 3 x - \cos 7 x)}{\sqrt{\sin 2 x + 1} - \sqrt{\tan 2 x + 1}} ?

(4) \lim_{x \to 0} \frac{\sin 2 x}{3 - \sqrt{3 x + 9}} ?

Answered by bobhans last updated on 05/Sep/20

solution :

(1) \lim_{x \to \frac{π}{4}} \sin (x - \frac{π}{4}) . \tan (x + \frac{π}{4}) =

set x = \frac{π}{4} + z; z \to 0

\lim_{z \to 0} \sin z . \tan (z + \frac{π}{2}) = \lim_{z \to 0} \frac{\sin z}{\cot (z + \frac{π}{2})}

\lim_{z \to 0} \frac{\sin z}{\tan z} = 1

(2) \lim_{x \to \frac{π}{2}} \frac{π (π - 2 x) \tan (x - \frac{π}{2})}{2 (x - π) \cos^{2} x} =

set x = \frac{π}{2} + s, s \to 0

\lim_{s \to 0} \frac{π (π - π - 2 s) . \tan s}{2 (s - \frac{π}{2}) \cos^{2} (s + \frac{π}{2})} =

\lim_{s \to 0} \frac{- 2 π s . \tan s}{(2 s - π) \sin^{2} s} = \frac{- 2 π}{- π} = 2

(3) \lim_{x \to 0} \frac{3 x (\cos 3 x - \cos 7 x)}{\sqrt{1 + \sin 2 x} - \sqrt{1 + \tan 2 x}} =

\lim_{x \to 0} \frac{3 x (1 - \frac{{9 x}^{2}}{2} - 1 + \frac{{49 x}^{2}}{2})}{(1 + \frac{\sin 2 x}{2}) - (1 + \frac{\tan 2 x}{2})} =

\lim_{x \to 0} \frac{2 \times 3 x ({20 x}^{2})}{\tan 2 x (\cos 2 x - 1)} = \lim_{x \to 0} \frac{{60 x}^{2}}{1 - {2 x}^{2} - 1}

= - 30

(4) \lim_{x \to 0} \frac{\sin 2 x}{3 - \sqrt{3 x + 9}} = \lim_{x \to 0} \frac{\sin 2 x}{3 - \sqrt{9 (1 + \frac{x}{3})}}

\lim_{x \to 0} \frac{\sin 2 x}{3 (1 - \sqrt{1 + \frac{x}{3}})} = \lim_{x \to 0} \frac{\sin 2 x}{3 (1 - (1 + \frac{x}{6}))}

= \lim_{x \to 0} \frac{\sin 2 x}{- (\frac{x}{2})} = - 4

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