How to calculate the limit? $$\lim_{x\rightarrow 0} (\sin(2\phi x)-2\phi x)\cot(\phi x)\csc^2(\phi x)=-\frac{4}{3}$$
where $\displaystyle \phi$ is a real number.
How to calculate the limit $\displaystyle \lim_{x\rightarrow 0} (\sin(2\phi x)-2\phi x)\cot(\phi x)\csc^2(\phi x)$
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calculus
proof-verification
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0I tried L'Hospital rule, but i don't have success! – 2017-01-16
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0Did you mean to say $\sin$? – 2017-01-16
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1@SimpleArt I believe $\operatorname{sen}$ is the Spanish for $\sin$ – 2017-01-16
1 Answers
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First see that
$$\sin(x)=x-\frac16x^3+\mathcal O(x^5)$$
So,
$$\sin(2\phi x)-2\phi x=\color{#4488dd}{-\frac43}\phi^3x^3+\mathcal O(x^5)$$
Similarly,
$$\cot(\phi x)\csc^2(\phi x)=\frac{\cos(\phi x)}{\sin^3(\phi x)}=\frac{\cos(\phi x)}{\phi^3x^3-2\phi^5x^5+\mathcal O(x^7)}$$
And combining all of this,
$$\begin{align}(\sin(2\phi x)-2\phi x)\cot(\phi x)\csc^2(\phi x)&=\frac{\cos(\phi x)(-\frac43\phi^3x^3+\mathcal O(x^5))}{\phi^3x^3-2\phi^5x^5+\mathcal O(x^7)}\\&=\frac{\cos(\phi x)(-\frac43+\mathcal O(x^2))}{1-2\phi^2x^2+\mathcal O(x^5)}\\&\to\color{#4488dd}{-\frac43}\end{align}$$