Graph the function f(x)= x sin( 1/x) and the equations y=|x| and y=-|x| in the same viewing window on your calculator. Then use the squeeze therom to find the limit of f(x) as it approaches zero.
I graphed the function and equations, but I am confused on how to find the limit using the squeeze therom.
If you look at the graphs at
http://www.wolframalpha.com/input/?i=plot+x*sin%281%2Fx%29%2C|x|%2C-|x|
it's easy to see the squeeze.
for x>=0, x = |x| and |sin(1/x)| <= 1
so, |x sin(1/x)| <= |x|
x sin(1/x) is squeezed in between the |x| lines. So, while f(0) does not exist, it is between |0| and -|0|, or just zero.
To find the limit of the function f(x) = x sin(1/x) as it approaches zero using the squeeze theorem, we need to establish upper and lower bounds for f(x) that converge as x approaches zero.
First, let's recall the squeeze theorem. The squeeze theorem states that if we have three functions, g(x), f(x), and h(x), such that g(x) ≤ f(x) ≤ h(x) for all x in the neighborhood of a particular point (except possibly at the point itself), and if lim(g(x)) = lim(h(x)) = L as x approaches that point, then lim(f(x)) = L as well.
In our case, we can use the squeeze theorem by finding two simpler functions that bound f(x) and have known limits as x approaches zero.
Let's find two functions, g(x) and h(x), that satisfy the conditions of the squeeze theorem.
For the lower bound, we can use the function g(x) = -|x|. Since |x| is always greater than or equal to -|x|, we have -|x| ≤ f(x) for all x. As x approaches zero, the limit of g(x) is also zero, so lim(g(x)) = 0 as x approaches zero.
For the upper bound, we can use the function h(x) = |x|. Since |x| is always greater than or equal to -|x|, we can still say that f(x) ≤ h(x) for all x. As x approaches zero, the limit of h(x) is also zero, so lim(h(x)) = 0 as x approaches zero.
Now, since both g(x) and h(x) have a limit of zero as x approaches zero, we can conclude that f(x) also has a limit of zero as x approaches zero by the squeeze theorem.