lim_(x→(π/2)) (((√(2tan^2 x+3))−(√(2tan^2 x+10)))/(cot x)) =?


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Question Number 163966 by bobhans last updated on 12/Jan/22

$\lim_{x \to \frac{π}{2}} \frac{\sqrt{2 \tan^{2} x + 3} - \sqrt{2 \tan^{2} x + 10}}{\cot x} = ?$
	Answered by cortano1 last updated on 12/Jan/22
	$set x=(π/2)+h⇒ { ((tan x=−cot h)),((cot x=−tan h)) :} lim_(h→0) (((√(2cot^2 h+3))−(√(2cot^2 h+10)))/(−tan h)) = lim_(h→0) (((√(2+3tan^2 h))−(√(2+10tan^2 h)))/(−tan^2 h)) = (√2) ×lim_(h→0) (((√(1+((3tan^2 h)/2)))−(√(1+5tan^2 h)))/(−tan^2 h)) =−(√2) ×lim_(h→0) ((((3/4)−(5/2))tan^2 h)/(tan^2 h)) =−(√2) ×(−(7/4))=((7(√2))/4)$
	$s e t x = \frac{π}{2} + h \Rightarrow {\begin{cases} \tan x = - \cot h \\ \cot x = - \tan h \end{cases}$ $\lim_{h \to 0} \frac{\sqrt{2 \cot^{2} h + 3} - \sqrt{2 \cot^{2} h + 10}}{- \tan h}$ $= \lim_{h \to 0} \frac{\sqrt{2 + 3 \tan^{2} h} - \sqrt{2 + 10 \tan^{2} h}}{- \tan^{2} h}$ $= \sqrt{2} \times \lim_{h \to 0} \frac{\sqrt{1 + \frac{3 \tan^{2} h}{2}} - \sqrt{1 + 5 \tan^{2} h}}{- \tan^{2} h}$ $= - \sqrt{2} \times \lim_{h \to 0} \frac{(\frac{3}{4} - \frac{5}{2}) \tan^{2} h}{\tan^{2} h}$ $= - \sqrt{2} \times (- \frac{7}{4}) = \frac{7 \sqrt{2}}{4}$
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