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Question Number 121031 by bramlexs22 last updated on 05/Nov/20

$$\underset{x\to 0}{\lim }\frac{\mathrm{x}+\mid \mathrm{x}\mid (1+\mathrm{x})}{\mathrm{x}}.\sin \left(\frac{1}{\mathrm{x}}\right)?$$

Commented by liberty last updated on 05/Nov/20

$$\underset{x\to 0^{-}}{\lim }\frac{\mathrm{x}+\mid \mathrm{x}\mid (1+\mathrm{x})}{\mathrm{x}}.\sin \left(\frac{1}{\mathrm{x}}\right)=$$$$\underset{x\to 0^{-}}{\lim }\frac{\mathrm{x}-\mathrm{x}-{\mathrm{x}}^2}{\mathrm{x}}.\sin \left(\frac{1}{\mathrm{x}}\right)=\underset{x\to 0^{-}}{\lim }-\mathrm{x}.\sin \left(\frac{1}{\mathrm{x}}\right)=0$$$$\mathrm{but}\underset{x\to 0^{+}}{\lim }\frac{\mathrm{x}+\mid \mathrm{x}\mid (1+\mathrm{x})}{\mathrm{x}}.\sin \left(\frac{1}{\mathrm{x}}\right)=$$$$\underset{x\to 0^{+}}{\lim }\frac{2\mathrm{x}+{\mathrm{x}}^2}{\mathrm{x}}.\sin \left(\frac{1}{\mathrm{x}}\right)=\underset{x\to 0^{+}}{\lim }(2+\mathrm{x}).\sin \left(\frac{1}{\mathrm{x}}\right)$$$$\mathrm{which}\mathrm{takes}\mathrm{on}\mathrm{every}\mathrm{value}\mathrm{between}-2\mathrm{and}2$$$$\mathrm{in}\mathrm{every}\mathrm{interval}(0,\delta ).\mathrm{Hence}\mathrm{g}\left(\mathrm{x}\right)=(2+\mathrm{x}).\sin \left(\frac{1}{\mathrm{x}}\right)$$$$\mathrm{does}\mathrm{not}\mathrm{approach}\mathrm{a}\mathrm{right}-\mathrm{hand}\mathrm{limit}$$$$\mathrm{at}\mathrm{x}\to 0\mathrm{from}\mathrm{the}\mathrm{right}.\mathrm{This}\mathrm{shows}$$$$\mathrm{that}\mathrm{a}\mathrm{function}\mathrm{may}\mathrm{have}\mathrm{a}\mathrm{limit}\mathrm{from}$$$$\mathrm{one}\mathrm{side}\mathrm{at}\mathrm{a}\mathrm{point}\mathrm{but}\mathrm{fail}\mathrm{to}\mathrm{have}\mathrm{a}\mathrm{limit}$$$$\mathrm{from}\mathrm{the}\mathrm{other}\mathrm{side}.▴$$

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