$$\lim_{z\rightarrow 0}\frac{12z^2+6\sin^2z-18(\cos z \sin z)z} {\sin^4z}$$
where $z$ goes to $0$.
Have been eyeballing it for quite a while, but without any luck. Any suggestions?
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1What have you tried? What techniques are available to you? Do you know L'Hopital's Rule? – Robert Shore Nov 04 '19 at 03:28
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@RobertShore oops, totally forgot about the L'HOpital's rule... I'm technically in the complex analysis now, so it slipped my mind. This should be quite easy now... I tried combinign terms, and adding and substracting, but whitout any luck. i wonder if you can solve it without the l'hopital's rule – Nick The Dick Nov 04 '19 at 03:30
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Use series expansion. – Pythagoras Nov 04 '19 at 03:46
2 Answers
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Use $\sin x = x - \frac16x^3+O(x^5)$ and expand the numerator to order $z^4$, matching the leading order of the denominator,
$$12z^2+6\sin^2z-18(\cos z \sin z)z$$ $$=12z^2+6(z-\frac16z^3)^2-9z\sin2z+O(z^6)$$ $$=12z^2+6z^2-2z^4-9z\left(2z-\frac16(2z)^3\right)+O(z^6)$$ $$=-2z^4+12z^4+O(z^6)$$ $$=10z^4+O(z^6)$$
Thus,
$$\lim_{z\rightarrow 0}\frac{12z^2+6\sin^2z-18(\cos z \sin z)z} {\sin^4z} =\lim_{z\rightarrow 0}\frac{10z^4} {z^4}=10$$
Quanto
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Using Are all limits solvable without L'Hôpital Rule or Series Expansion
$$\dfrac{y^2-\sin^2y}{y^4}=\dfrac{1+1}{3!}$$
Now observe that $$12z^2+6\sin^2z-18z(\cos z\sin z)=-6(z^2-\sin^2z)+9z(2z-\sin2z)$$
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