15th September 2006 - 06:53 PM
How can you tell if f(x) has a limit as x approaches c?
I just started Calc 1, and I'm having a little trouble interpreting graphs, and determining if the function has a limit. Specifically, let's say you have the function y=x. Now let's say that the point (2,2) is NOT included in the function. (so far, so good; as x approaches 2, lim f(x) = 2). Now, let's say that in addition to this line, you have the point (2,10) INCLUDED in the function (a "piecewise" function). Here's where I get confused. Basically the function when graphed looks like a line with one NOT-included point on it, and then somewhere above that point (same x coordinate) there is another point that is included in the function. Do these sort of graphs always have the same limits as a function that does NOT include the extra point above the line (again, the point above the line has the same x coordinate as the NOT-included point)? Does this extra point affect the function's limit in any way?
Thank you very much!
15th September 2006 - 10:33 PM
Yes, the limit is the value of what would have been in the hole in the function. In your example, the limit as x approaches 2 would be 2. Limits look at what value a function approaches, not what the actual value is. So in the example, although f(2)=10, the function is approaching 2. Remember that f(x) has to approach the same value from both the left and the right of the point.
I'm sure you have or will learn these soon, but they are good to know:
Conditions under which there are no limits:
2. Left and right sides approach different numbers
3. Continuous Oscillation
There might be a fourth, but I can't remember
15th September 2006 - 11:19 PM
If f(x) = 1 if x is rational, and f(x) = 0 in all other cases, then f(x) is everywhere discontinuous and has no limit, even though we can prove that f is a well-defined function and specifically that f(1)=1 and f(pi)=0. This is the fourth case of Amac, the case of the discontinuous graph.
sin(1/x) has no limit as x->0
x sin(1/x) has a limit as x->0, which you can prove with epsilon-delta method.
sin(x)/x has a limit as x->0, which is simple to see if you graph it, but I think you need calculus to prove it is 1. (for example, Calculus lets you substitute a Taylor series for sin(x), so that sin(x)/x = 1 + A x^2 + B x^4 + C x^6 ... which converges everywhere.)