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What will be the solution of expression

lim(x approches to 0) sin(1/x)

The value of Sin 1/x will fluctuate between -1 and +1. When the measure of angle represented by 1/x i in degrees, starting from 1/x = 0, Sign1/x is equal to 0. It increase to 1 when angle equals 90 degrees, again falling to 0 when angle equals 180 degrees. When 1/x becomes 270 degrees Sin 1/x is -1. When 1/x increase to 360 degrees the cycle is complete and Sin 1/x, again becomes 1. This cycle is repeated for every 360 degrees increase in value of 1/x.

As value of x is reduced the value of 1/x will go on increasing, and the cycle of 360 degrees for 1/x will keep on repeating. With this value of Sin 1/x will keep in fluctuating between -1 and +1, never approaching any fixed value.

The more x tends to 0, the more (1/x) increases limitless. So, it doesn't matter how the interval centered in 0 is chosen, in any of all these intervals, there will be values of x, so that the sine of these values, will be -1, 0 or 1.

The conclusion would be that sin(1/x) is oscillating etween the maximum value,1, and the minimum value,(-1), in any neighbourhood chosen around the 0 value.

So, lim sin(1/x), does not exist!