evaluate the limit as x approaches 0 from the right. of the function xcos(1/x)
I know the answer is O but I am not sure as of my reasoning. I just that that since cos(1/x) is being multiplied by a number approching 0 that the whole limit would be o.
Let's just use an intuitive approach.
We know that if x -->0 , 1/x becomes infinitely large.
But cos(anything) is always a number between -1 and +1, no matter how large 1/x becomes
So we are then multiplying this by "almost zero" , so the answer will be zero.
more formally, for the purposes of homework, just use the comparison test
|cos(1/x)| <= 1, so
x cos(1/x) < x
which -> 0
To evaluate the limit as x approaches 0 from the right of the function f(x) = xcos(1/x), we need to consider the behavior of the function as x approaches 0 from the right.
We can start by simplifying the equation:
f(x) = x * cos(1/x)
As x approaches 0 from the right, the value of 1/x becomes infinitely large. In this case, the cosine function does not converge to a specific value but oscillates between -1 and 1.
However, the function x multiplied by the cosine function approaches 0 as x approaches 0 from the right. This is because the function x approaches 0 faster than the oscillations of the cosine function can affect it.
To show this more formally, we can use the Squeeze Theorem, also known as the Sandwich Theorem:
-1 ≤ cos(1/x) ≤ 1
Multiply all three sides of the inequality by x:
-|x| ≤ x * cos(1/x) ≤ |x|
Since the limit as x approaches 0 from the right of both the lower and upper bounds is 0, we can conclude that:
lim(x→0+) x * cos(1/x) = 0
Therefore, the limit as x approaches 0 from the right of the function xcos(1/x) is indeed 0.