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Question Number 130925 by Chhing last updated on 30/Jan/21

Problem

Without L^{'} Hopital

calculate

\lim_{x \to 0} \frac{\tan^{2} (x) - x^{2} \cos (2 x)}{x^{2} - \sin^{2} (x)}

Commented by malwan last updated on 30/Jan/21

x \to 0

\Rightarrow t a n x = s i n x \land c o s 2 x = 1

\underset{x \to 0}{l i m} \frac{{s i n}^{2} x - x^{2}}{x^{2} - {s i n}^{2} x} = - 1

I t i s n o t r i g h t

w h a y ?

Commented by Chhing last updated on 30/Jan/21

but \tan (0) = 0

Commented by malwan last updated on 30/Jan/21

a n d s i n (0) = 0

Commented by Chhing last updated on 31/Jan/21

Yes, sir

Answered by bramlexs22 last updated on 30/Jan/21

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

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

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

= \lim_{x \to 0} \frac{(\frac{8 x^{2}}{3})}{(\frac{x^{2}}{3})}

= 8 ✓

Commented by bramlexs22 last updated on 30/Jan/21

Answered by benjo_mathlover last updated on 30/Jan/21

\lim_{x \to 0} \frac{\tan^{2} x - x^{2} (\cos^{2} x - \sin^{2} x)}{x^{2} - \sin^{2} x}

\lim_{x \to 0} \frac{\tan^{2} x - x^{2} \cos^{2} x + x^{2} \sin^{2} x}{(x + \sin x) (x - \sin x)}

\lim_{x \to 0} \frac{(\tan x + xcos x) (\tan x - xcos x) + x^{2} \sin^{2} x}{(x + \sin x) (x - \sin x)}

\lim_{x \to 0} \frac{(\frac{\tan x}{x} + \cos x) (x + \frac{x^{3}}{3} - x (1 - \frac{1}{2} x^{2})) + x^{2} (x - \frac{x^{3}}{6})}{(1 + \frac{\sin x}{x}) (x - x + \frac{x^{3}}{6})}

\lim_{x \to 0} \frac{2 (\frac{x^{3}}{3} + \frac{x^{3}}{2}) + x^{3} - \frac{x^{5}}{6}}{2 (\frac{x^{3}}{6})} = \lim_{x \to 0} \frac{\frac{8}{3} - \frac{x^{2}}{6}}{\frac{1}{3}}

= 8 .

Commented by bramlexs22 last updated on 30/Jan/21

y e a h h h \dots ✓

Commented by Chhing last updated on 30/Jan/21

Thank you

Commented by Chhing last updated on 30/Jan/21

Thank you

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