Find the limit of cos⁡(3x)−1sin⁡(2x)tan⁡(3x) without L'Hospital technique

limx→0cos⁡3x−1sin⁡2xtan⁡3x=−limx→0sin23xcos⁡3xsin⁡2xsin⁡3x(1+cos⁡3x)=−32⋅limx→0sin⁡3x3x⋅1limx→0sin⁡2x2x⋅limx→0cos⁡3x1+cos⁡3x=?

limx→0cos(3x)-1sin(2x)tan(3x)=limx→0-2sin2(3x2)cos(3x)sin(2x)sin(3x)=−2limx→0sin2(3x2)(3x2)2(2x)sin⁡(2x)(3x)(sin⁡(3x)}.(3x2)2(2x)(3x)cos⁡(3x)=−34

limx→0cos⁡3x−1sin⁡2xtan⁡3x=−limx→0sin23xcos⁡3xsin⁡2xsin⁡3x(1+cos⁡3x)=−32⋅limx→0sin⁡3x3x⋅1limx→0sin⁡2x2x⋅limx→0cos⁡3x1+cos⁡3x=?

limx→0cos(3x)-1sin(2x)tan(3x)=limx→0-2sin2(3x2)cos(3x)sin(2x)sin(3x)=−2limx→0sin2(3x2)(3x2)2(2x)sin⁡(2x)(3x)(sin⁡(3x)}.(3x2)2(2x)(3x)cos⁡(3x)=−34

Expert answers to "Find the limit of cos⁡(3x)−1sin⁡(2x)tan⁡(3x) without L'Hospital technique"

Colten Welch

Answered question

2022-04-13

Find the limit of

\frac{\cos (3 x) - 1}{\sin (2 x) \tan (3 x)}

without L'Hospital technique

Answer & Explanation

webhui2v2

Beginner2022-04-14Added 10 answers

lim_{x \to 0} \frac{\cos 3 x - 1}{\sin 2 x \tan 3 x} = - lim_{x \to 0} \frac{\sin^{23} x \cos 3 x}{\sin 2 x \sin 3 x (1 + \cos 3 x)}

= - \frac{3}{2} \cdot lim_{x \to 0} \frac{\sin 3 x}{3 x} \cdot \frac{1}{lim_{x \to 0} \frac{\sin 2 x}{2 x}} \cdot lim_{x \to 0} \frac{\cos 3 x}{1 + \cos 3 x} = ?

shanna87mn

Beginner2022-04-15Added 19 answers

$\lim_{x \to 0} \frac{\cos (3 x) - 1}{\sin (2 x) \tan (3 x)} = \lim_{x \to 0} \frac{- 2 \sin^{2} (\frac{3 x}{2}) \cos (3 x)}{\sin (2 x) \sin (3 x)}$
$= - 2 lim_{x \to 0} \frac{\sin^{2} (\frac{3 x}{2})}{{(\frac{3 x}{2})}^{2}} \frac{(2 x)}{\sin (2 x)} \frac{(3 x)}{(\sin (3 x)} . \frac{{(\frac{3 x}{2})}^{2}}{(2 x) (3 x)} \cos (3 x)}$
$= - \frac{3}{4}$

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