calculate lim_(x→0) ((ln(1+tan(2x))−ln(cos(3x)))/x^3 )


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Question Number 62657 by mathmax by abdo last updated on 24/Jun/19

$c a l c u l a t e {l i m}_{x \to 0} \frac{l n (1 + t a n (2 x)) - l n (c o s (3 x))}{x^{3}}$
Commented by mathmax by abdo last updated on 24/Jun/19
$let use hospital teorem with u(x)=ln(1+tan(2x))−ln(cos(3x)) and v(x) =x^3 we have u^′ (x) =((2(1+tan^2 (2x)))/(1+tan(2x)))+ ((3sin(3x))/(cos(3x))) =2 (((1+tan(2x))^2 −2tan(2x))/(1+tan(2x))) +3 tan(3x) =2(1+tan(2x))−4 ((tan(2x)+1−1)/(1+tan(2x))) +3tan(3x) =2(1+tan(2x))−4 +(4/(1+tan(2x))) +3 tan(3x) =−2 +2tan(2x) +3tan(3x)+(4/(1+tan(2x))) ⇒ u^((2)) (x) =4(1+tan^2 (2x)) +9(1+tan^2 (3x))+4((−2(1+tan^2 (2x)))/((1+tan(2x))^2 )) ⇒ u^((3)) (x) =4{4tan(2x)(1+tan^2 (2x))}+ 9{ 6tan(3x)(1+tan^2 (3x)} −8((4tan(2x)(1+tan^2 (2x)(1+tan(2x))^2 −(1+tan^2 (2x))2(1+tan(2x))2(1+tan^2 (2x)))/((1+tan(2x))^4 )) ⇒lim_(x→0) u^((3)) (x) =−8 ((−4)/1) =32 also we have v^′ (x)=3x^2 ⇒v^((2)) (x)=6x ⇒v^((3)) (x)=6 ⇒lim_(x→0) ((ln(1+tan(2x))−ln(cos(3x)))/x^3 ) =((32)/6) =((16)/3) .$
$l e t u s e h o s p i t a l t e o r e m w i t h u (x) = l n (1 + t a n (2 x)) - l n (c o s (3 x))$ $a n d v (x) = x^{3} w e h a v e$ $u^{^{'}} (x) = \frac{2 (1 + {t a n}^{2} (2 x))}{1 + t a n (2 x)} + \frac{3 s i n (3 x)}{c o s (3 x)}$ $= 2 \frac{{(1 + t a n (2 x))}^{2} - 2 t a n (2 x)}{1 + t a n (2 x)} + 3 t a n (3 x)$ $= 2 (1 + t a n (2 x)) - 4 \frac{t a n (2 x) + 1 - 1}{1 + t a n (2 x)} + 3 t a n (3 x)$ $= 2 (1 + t a n (2 x)) - 4 + \frac{4}{1 + t a n (2 x)} + 3 t a n (3 x)$ $= - 2 + 2 t a n (2 x) + 3 t a n (3 x) + \frac{4}{1 + t a n (2 x)} \Rightarrow$ $u^{(2)} (x) = 4 (1 + {t a n}^{2} (2 x)) + 9 (1 + {t a n}^{2} (3 x)) + 4 \frac{- 2 (1 + {t a n}^{2} (2 x))}{{(1 + t a n (2 x))}^{2}} \Rightarrow$ $u^{(3)} (x) = 4 {4 t a n (2 x) (1 + {t a n}^{2} (2 x))} + 9 {6 t a n (3 x) (1 + {t a n}^{2} (3 x)}$ $- 8 \frac{4 t a n (2 x) (1 + {t a n}^{2} (2 x) {(1 + t a n (2 x))}^{2} - (1 + {t a n}^{2} (2 x)) 2 (1 + t a n (2 x)) 2 (1 + {t a n}^{2} (2 x))}{{(1 + t a n (2 x))}^{4}}$ $\Rightarrow {l i m}_{x \to 0} u^{(3)} (x) = - 8 \frac{- 4}{1} = 32 a l s o w e h a v e v^{^{'}} (x) = 3 x^{2} \Rightarrow v^{(2)} (x) = 6 x \Rightarrow v^{(3)} (x) = 6$ $\Rightarrow {l i m}_{x \to 0} \frac{l n (1 + t a n (2 x)) - l n (c o s (3 x))}{x^{3}} = \frac{32}{6} = \frac{16}{3} .$
Commented by MJS last updated on 24/Jun/19

$I {don}^{'} t think we can use l^{'} Hopital a second$ $time because u^{'} (0) = 2$
	Answered by MJS last updated on 24/Jun/19

	$\lim_{x \to 0} \frac{\ln (1 + \tan 2 x) - \ln \cos 3 x}{x^{3}} =$ $= \lim_{x \to 0} \frac{\frac{d}{d x} [\ln (1 + \tan 2 x) - \ln \cos 3 x]}{\frac{d}{d x} [x^{3}]} =$ $= \lim_{x \to 0} \frac{\frac{4}{1 + \cos 4 x + \sin 4 x} + 3 \tan 3 x}{3 x^{2}} = \frac{2}{0} \Rightarrow undefined$
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