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Question Number 170141 by sciencestudent last updated on 17/May/22

$$li\underset{x\to 0}{m}(\frac{1}{{sin}^{2}x}-\frac{1}{x^2})=?$$

Commented by mr W last updated on 17/May/22

$$=(\frac{1}{\sin x}+\frac{1}{x})(\frac{1}{\sin x}-\frac{1}{x})$$$$=(\frac{x}{\sin x}+1)\left(\frac{x-\sin x}{x^3}\right)\frac{x}{\sin x}$$$$=(\frac{x}{\sin x}+1)\left(\frac{x-x+\frac{x^3}{3!}-\frac{x^5}{5!}+...}{x^3}\right)\frac{x}{\sin x}$$$$=(\frac{x}{\sin x}+1)(\frac{1}{3!}-\frac{x^2}{5!}+...)\frac{x}{\sin x}$$$$=(1+1)\times \left(\frac{1}{3!}\right)\times 1$$$$=\frac{2}{3!}$$$$=\frac{1}{3}$$

Commented by sciencestudent last updated on 17/May/22

$$HowdidYousimplipy(\frac{1}{sinx}-\frac{1}{x})(\frac{1}{sinx}+\frac{1}{x})?$$

Commented by mr W last updated on 17/May/22

$$a^2-b^2=(a-b)(a+b)$$

Answered by ajfour last updated on 17/May/22

$$l=\underset{x\to 0}{\lim }(\frac{2}{1-\cos 2x}-\frac{1}{x^2})$$$$=\underset{x\to 0}{\lim }(\frac{2}{1-[1-\frac{4x^2}{2!}+\frac{16x^4}{4!}-..]}-\frac{1}{x^2})$$$$=\underset{x\to 0}{\lim }\frac{1}{x^2}(\frac{}{1-\frac{1x^2}{3}+{sx}^4-..}-1)$$$$=\underset{x\to 0}{\lim }\left(\frac{\frac{1}{3}-{sx}^2+...}{1-\frac{x^2}{3}+{sx}^4-...}\right)=\frac{1}{3}$$

Commented by sciencestudent last updated on 17/May/22

$$Howtoget1-\frac{x^2}{3}+{sx}^4-\cdot \cdot \cdot ?$$

Commented by mr W last updated on 17/May/22

$$\sin \left(x\right)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$$$$\cos \left(x\right)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...$$$$readmoreaboutMaclaurinseries...$$

Commented by mr W last updated on 17/May/22

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