quietrain said:
the lim[]it of cos(1/x) , x-->0, is = cos(huge number) is not equal to cos(1/0) = undefined
f(0) = 0 as defined in f.
f:= \left\{ \begin{array}{ll} \cos\left( \frac{1}{x} \right) & \mbox{, if } x \neq 0 \\ 0 & \mbox{, if } x =0 \end{array} \right.
so it is discontinuous at x=0 right?
Yes, that's true but that's not how you prove it. You have to express your idea using Mathematical terms, like this:
Since f(0) is defined to be 0. And we know that the function is not continuous at 0. This is because cos(x) has the period of 2 \pi. However, when x \rightarrow 0, \frac{1}{x} \rightarrow \infty, which means that as
x tends to 0, \cos \left( \frac{1}{x} \right) oscillates between -1, and 1. The
closer x gets to 0, the
faster \frac{1}{x} grows without bounds, this leads to the fact that \cos \left( \frac{1}{x} \right) oscillates
faster.
This is how you 'imagine'. But you have to write your idea down Mathematically. Like this:
- Choose a sequence xn that tends to 0.
- But f(xn) does not tend to f(0) = 0 (as defined)
We choose the sequence x_n = \frac{1}{2n \pi}. Of course this sequence tends to 0, as
n grows without bound, right? But we have:
\begin{align}<br />
f(x_n) &= \cos\left( \frac{1}{\x_n} \right) \mbox{ since } x_n \neq 0, \forall n \\<br />
&= \cos(2n \pi)\\<br />
&= 1 \rightarrow 1 \neq f(0) = 0<br />
\end{align}
similarly, for (y-5)cos(1/x2)
the limit x,y --> 0,y is the same as above, so it is discontinous at x=0 and y = any value except 5
Yup, but can you present this Mathematically?
because at 5,
limit x,y --> 0,5 = 0 which is = f(0,5) = 0
so it is continuous at 0,5
Well, yes. But can you prove that \lim_{(x, y) \rightarrow (0; 5)} f(x; y) = 0?
so my question is , is the only way to spot 0,5 through inspection or is there a fix method of finding such "special" points?
Yes, the more you practice, the better you'll get at spotting out these
special points. One method to spot out these point has been mentioned earlier in the post. Read it carefully, and see if you can get the point.
SammyS said:
You may also evaluate \textstyle \lim_{(x, y)\to(0, 5)}\,f(x,y) along an arbitrary direction. It's easy to do along the line x = 0 or y = 5. Both of these obviously give zero. To approach (0, 5) from an arbitrary direction, let y = m\,x+5 and evaluate \textstyle \lim_{x\to 0}\,f(x,m\,x+5).
No, this is not
arbitrary enough. (x; y) not only tends to (0; 5) in straight lines, it can also tends to (0; 5) in
many curves.
You can consider this example:
f(x; y) := \left\{ \begin{array}{ll} \frac{x ^ 3 y}{x ^ 6 + y ^ 2} & \mbox{ , if } (x; y) \neq (0; 0) \\ 0 & \mbox { , if } (x; y) = (0; 0) \end{array} \right.
If (x; y) tends to (0; 0) along straight lines, i.e y = mx.
We have:
f(x, mx) = \frac{mx ^ 4}{x ^ 6 + m ^ 2 x ^ 2} = \frac{mx ^ 2}{x ^ 4 + m ^ 2} \rightarrow 0
However, if we go along the curve y = x
3, we have:
f(x, x ^ 3) = \frac{x ^ 6}{x ^ 6 + x ^ 6} = \frac{1}{2} \rightarrow \frac{1}{2}.
