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Question Number 128187 by john_santu last updated on 05/Jan/21

$$\left(1\right)\underset{x\to 0}{\lim }\frac{x}{\sqrt{1-\cos x}}=?$$$$\left(2\right)\underset{x\to 0}{\lim }\frac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x-\sin \beta x}=?$$

Answered by Dwaipayan Shikari last updated on 05/Jan/21

$$\underset{x\to 0}{\lim }\frac{e^{ax}-e^{\beta x}}{sin\alpha x-sin\beta x}=\frac{1+\alpha x-1-\beta x}{\alpha x-\beta x}=1\underset{x\to 0}{\lim }{e}^x=1+xandsinx=x$$

Answered by liberty last updated on 05/Jan/21

$$\left(1\right)\underset{x\to 0^{-}}{\lim }\frac{\mathrm{x}}{\sqrt{2\sin {}^2\left(\frac{1}{2}\mathrm{x}\right)}}=\underset{x\to 0^{-}}{\lim }\frac{1}{\sqrt{2}}\frac{\mathrm{x}}{\mid \sin \frac{\mathrm{x}}{2}\mid }$$$$=\frac{1}{\sqrt{2}}.\underset{x\to 0^{-}}{\lim }\frac{\mathrm{x}}{-\sin \frac{\mathrm{x}}{2}}=-\frac{2}{\sqrt{2}}$$$$\underset{x\to 0^{+}}{\lim }\frac{\mathrm{x}}{\sqrt{2}\mid \sin \frac{\mathrm{x}}{2}\mid }=\frac{2}{\sqrt{2}}$$$$\mathrm{the}\mathrm{limit}\mathrm{does}\mathrm{not}\mathrm{exist}$$

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