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Alex
3 years ago
11

Lim x-0 (sin2xcsc3xsec2x)/x²cot²4x

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
8 0

By the definitions of cosecant, secant, and cotangent, we have

\dfrac{\sin2x\csc3x\sec2x}{x^2\cot^24x}=\dfrac{\sin2x\sin^24x}{x^2\sin3x\cos2x\cos^24x}

Then we rewrite the fraction as

\dfrac{\sin2x}{2x}\left(\dfrac{\sin4x}{4x}\right)^2\dfrac{3x}{\sin3x}\dfrac{32}{3\cos2x\cos^24x}

The reason for this is that we want to apply the well-known limit,

\displaystyle\lim_{x\to0}\frac{\sin ax}{ax}=\lim_{x\to0}\frac{ax}{\sin ax}=1

for a\neq0. So when we take the limit, we have

\displaystyle\lim_{x\to0}\cdots=\lim_{x\to0}\frac{\sin2x}{2x}\left(\lim_{x\to0}\frac{\sin4x}{4x}\right)^2\lim_{x\to0}\frac{3x}{\sin3x}\lim_{x\to0}\frac{32}3\cos2x\cos^24x}

=1\cdot1^2\cdot1\cdot\dfrac{32}3=\dfrac{32}3

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The data we have is:

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to solve it we must isolate x:

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