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Question Number 167699 by HongKing last updated on 23/Mar/22

Answered by alephzero last updated on 23/Mar/22

$$1.\underset{x\to 0}{\lim }\frac{1-\cos x}{x\sin x}\stackrel{\mathrm{use}{\mathrm{L}}^{\prime }\mathrm{H}{\widehat{\mathrm{o}pital}}^{\prime }\mathrm{s}\mathrm{rule}}{=}$$$$=\lim \frac{\sin x}{\sin x+x\cos x}\stackrel{\mathrm{again}}{=}$$$$=\lim \frac{\cos x}{\cos x-\sin x+\cos x}=$$$$=\lim \frac{\cos x}{2\cos x-\sin x}=\frac{1}{2\times 1-0}=\frac{1}{2}$$

Answered by alephzero last updated on 23/Mar/22

$$2.{\mathrm{L}}^{\prime }\mathrm{Hopital}\mathrm{again}$$$$\lim \frac{{3\cos }^{2}2x-\cos 2x-2}{x\cos x}=$$$$\lim \frac{-6\sin 4x+2\sin 2x}{\cos x-x\sin x}=$$$$\frac{0+0}{1-0}=0$$

Answered by qaz last updated on 23/Mar/22

$$\underset{\mathrm{x}\to 0}{\lim }\frac{({\mathrm{e}}^{-3\mathrm{x}+2}-{\mathrm{e}}^2)\sin \pi \mathrm{x}}{{4\mathrm{x}}^2}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^2\pi \mathrm{x}({\mathrm{e}}^{-3\mathrm{x}}-1)}{{4\mathrm{x}}^2}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\pi {\mathrm{e}}^2}{4\mathrm{x}}(-3\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right))$$$$=-\frac{3}{4}\pi {\mathrm{e}}^2$$

Answered by malwan last updated on 24/Mar/22

$$1.\underset{x\to 0}{lim}\frac{1-cosx}{xsinx}=\underset{x\to 0}{lim}\frac{2{sin}^2\frac{x}{2}}{xsinx}$$$$=2\times {\left(\frac{1}{2}\right)}^2=\frac{1}{2}$$

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