Step by step solution to "How to solve this limit:limx→0tan⁡6xsin⁡2x"

wibawanyacl3q

Answered question

2022-03-30

How to solve this limit:

lim_{x \to 0} \frac{\tan 6 x}{\sin 2 x}

Lesly Fernandez

Beginner2022-03-31Added 16 answers

Notice,

lim_{x \to 0} \frac{\tan 6 x}{\sin 2 x} = lim_{x \to 0} \frac{6 x (\frac{\tan 6 x}{6 x})}{2 x (\frac{\sin 2 x}{2 x})}

= 3 lim_{x \to 0} \frac{(\frac{\tan 6 x}{6 x})}{(\frac{\sin 2 x}{2 x})}

= 3 \cdot \frac{lim_{x \to 0} \frac{\tan 6 x}{6 x}}{lim_{x \to 0} \frac{\sin 2 x}{2 x}}

= 3 \cdot \frac{1}{1} = 3

Jonathon Hanson

Beginner2022-04-01Added 9 answers

lim_{x \to 0} \frac{\frac{\sin (6 x)}{\cos (6 x)}}{\sin (2 x)}

= lim_{x \to 0} \frac{\sin (6 x)}{\cos (6 x) \sin (2 x)}

= [lim_{x \to 0} \frac{1}{\cos (6 x)}] [lim_{x \to 0} \frac{\sin (6 x)}{\sin (2 x)}]

= [\frac{1}{\cos (6 \cdot 0)}] [lim_{x \to 0} \frac{\sin (6 x)}{\sin (2 x)}]

= [\frac{1}{1}] [lim_{x \to 0} \frac{\sin (6 x)}{\sin (2 x)}]

= [1] [lim_{x \to 0 \frac{\sin (6 x)}{\sin (2 x)}}]

= lim_{x \to 0} \frac{\sin (6 x)}{\sin (2 x)}

= lim_{x \to 0} \frac{6 \cos (6 x)}{2 \cos (2 x)}

= lim_{x \to 0} \frac{3 \cos (6 x)}{\cos (2 x)}

= \frac{3 \cos (6 \cdot 0)}{\cos (2 \cdot 0)} = \frac{3 \cos (0)}{\cos (0)} = \frac{3}{1} = 3

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