Plz help me out with a solution...
f(x) = sin(1/x)
when x is not = to 0 and f(0) = 0
Thanx in advance...
A function is continuous if
Limit of y-->x of f(y) equals f(x) for every x. In this case you have t see if this is the case for x = 0. f(0) = 0, therefore we need to calculate:
Limit x--->0 f(x)
It turns out that this limit does not exist, therefore the function is not continuous. However, for this problem you only need to show that the limit cannot be equal to zero.
Suppose the limit is zero, then for every interval I from minus delta to plus delta you can always find an epsilon such that the function maps the interval ranging from -epsilon to plus epsion into the interval I. So, you can make delta arbitrarily small and still find a range around zero that is mapped by f to I.
In this case this is clearly not true. The function oscillates between minus 1 and 1 for arbitrary small values of x. This means that for delta smaller than one, there is no interval containing zero which will be mapped by f to the range from minus delta to delta, because any interval containing zero, no matter how small, is mapped by f to an interval ranging from minus 1 to 1.
To show that the limit of f(x) as x approaches 0 does not exist, we can use the definition of limit.
Let's consider the left-hand limit first. As x approaches 0 from the negative side (x < 0), the function sin(1/x) oscillates between -1 and 1 infinitely many times. It does not converge to a single value.
Next, let's consider the right-hand limit. As x approaches 0 from the positive side (x > 0), the function sin(1/x) oscillates between -1 and 1 infinitely many times. Again, it does not converge to a single value.
Since the left-hand limit and the right-hand limit are not equal, we can conclude that the limit of f(x) as x approaches 0 does not exist.
Therefore, the function f(x) = sin(1/x) is not continuous at x = 0.