Showing that limx→0xsin⁡(1x)−cos⁡(1x)x does not exist without sequence Show that limx→0xsin⁡(1x)−cos⁡(1x)x does not exist.

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Showing that limx→0xsin⁡(1x)−cos⁡(1x)x does not exist without sequence  Show that  limx→0xsin⁡(1x)−cos⁡(1x)x  does not exist.

saennwegoyk · Accepted Answer

−1≤sin⁡(1x)≤1⇒−x≤xsin⁡(1x)≤x and −1≤cos⁡(1x)≤1  −1−x≤xsin⁡(1x)−cos⁡(1x)≤1+x⇒−1−xx≤xsin⁡(1x)−cos⁡(1x)x≤1+xx  We can show that as x→0 the limit is unbounded.

search633504 · Accepted Answer

Suppose it does exist. Then limx→0xsin⁡(1x)−cos⁡(1x)=0. Since limx→0xsin⁡(1x)=0 this would mean limx→0cos⁡(1x)=0 which is the same as: limy→∞cos⁡(y)=0. So the proof simplifies to showing this last limit is not zero.  One way of showing that limy→∞cos⁡(y) does not exist without using sequences would be to use this result (good exercise to try to prove):  if f:R→R is a periodic function and limx→∞f(x) exists. Then f is constant.

Showing that \lim_{x \to 0}\frac{x\sin(1/x)-\cos(1/x)}x does not exist without sequence Show

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