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Question Number 112851 by bemath last updated on 10/Sep/20

$$\mathrm{Without}{\mathrm{L}}^{\prime }\mathrm{Hopital}$$$$\underset{x\to 0}{\lim }(\frac{\mathrm{a}}{\mathrm{x}}-\cot \frac{\mathrm{x}}{\mathrm{a}})?$$

Answered by bobhans last updated on 10/Sep/20

$$\mathrm{set}\frac{\mathrm{x}}{\mathrm{a}}=\mathrm{t}\to \frac{\mathrm{a}}{\mathrm{x}}=\frac{1}{\mathrm{t}}$$$$\mathrm{L}=\underset{\mathrm{t}\to 0}{\lim }(\frac{1}{\mathrm{t}}-\cot \mathrm{t})=\underset{\mathrm{t}\to 0}{\lim }(\frac{1}{\mathrm{t}}-\frac{1}{\tan \mathrm{t}})$$$$\underset{\mathrm{t}\to 0}{\lim }\left(\frac{\tan \mathrm{t}-\mathrm{t}}{\mathrm{t}.\tan \mathrm{t}}\right).[\mathrm{let}\mathrm{t}=2\mathrm{w}]$$$$\mathrm{L}=\underset{\mathrm{w}\to 0}{\lim }\frac{\tan 2\mathrm{w}-2\mathrm{w}}{2\mathrm{w}.\tan 2\mathrm{w}}=\underset{\mathrm{w}\to 0}{\lim }\frac{\frac{2\tan \mathrm{w}}{1-{\tan }^2\mathrm{w}}-2\mathrm{w}}{2\mathrm{w}.\left(\frac{2\tan \mathrm{w}}{1-\tan {}^2\mathrm{w}}\right)}$$$$\mathrm{L}=\underset{\mathrm{w}\to 0}{\lim }\frac{2\tan \mathrm{w}-2\mathrm{w}+2\mathrm{wtan}{}^2\mathrm{w}}{4\mathrm{w}.\tan \mathrm{w}}$$$$\mathrm{L}=\underset{\mathrm{w}\to 0}{\lim }\frac{\tan \mathrm{w}-\mathrm{w}}{2\mathrm{w}.\tan \mathrm{w}}+\underset{\mathrm{w}\to 0}{\lim }\frac{2\mathrm{wtan}{}^2\mathrm{w}}{4\mathrm{wtan}\mathrm{w}}$$$$\mathrm{L}=\frac{1}{2}\mathrm{L}+0\Rightarrow \mathrm{L}=0$$

Answered by Dwaipayan Shikari last updated on 10/Sep/20

$$\frac{a}{x}-cot\frac{x}{a}=\frac{\frac{a}{x}tan\frac{x}{a}-1}{tan\frac{x}{a}}=\frac{\frac{a}{x}\frac{sin\frac{x}{a}}{cos\frac{x}{a}}-1}{\frac{x}{a}}=\frac{a}{x}.\frac{1-cos\frac{x}{a}}{cos\frac{x}{a}}$$$$=\frac{a}{x}.\frac{2{sin}^2\frac{x}{a}}{1}=2.\frac{a}{x}.\frac{x^2}{a^2}=\frac{2x}{a}=0$$

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