For what value of the constant c is the function f continuous on (−∞,∞) where
f(x)= c if x=0, and
xsin(1/x) otherwise
since |sin(1/x)|<=1 x*sin(1/x) -> 0 as x->0
f(0)=0
will make f defined everywhere.
To determine the value of the constant c that makes the function f continuous on the interval (-∞, ∞), we need to evaluate the limit of f(x) as x approaches 0 from both the left and right sides.
First, let's consider the left-hand limit as x approaches 0:
lim(x->0-) f(x) = lim(x->0-) xsin(1/x)
Since sin(1/x) is bounded between -1 and 1, the product xsin(1/x) will approach 0 as x approaches 0. Therefore, we can say that:
lim(x->0-) f(x) = 0
Next, let's consider the right-hand limit as x approaches 0:
lim(x->0+) f(x) = lim(x->0+) xsin(1/x)
In this case, the function xsin(1/x) oscillates infinitely as x gets closer to 0. The limit of this function as x approaches 0 does not exist.
For f(x) to be continuous at x = 0, the left-hand limit and the right-hand limit must be equal. Since the right-hand limit does not exist, for f(x) to be continuous on (-∞, ∞), the only possible value for c is:
c = 0
By setting c = 0, the function f(x) becomes:
f(x) = 0 if x = 0,
f(x) = xsin(1/x) otherwise
Therefore, the constant c must be 0 for the function f(x) to be continuous on (-∞, ∞).