Undefined Points

[Mentallic]

I am aware that for a function that is undefined at a point x=a such as f(x)=1/(x-a)

\underbrace{lim}_{x\rightarrow a}f(x)=\pm \infty

But it tends to infinite only because it is in the form a/0, where a\neq0.

Undefined values in the form 0/0 can have a range of values - all reals if I'm not mistaken.

I thus set up a function f(x) multiplied by another function g(x) so that f(a)=0 and g(a) undefined. However, the functions are not in a form where they can seemingly cancel factors of the zero and undefined value.

e.g.
h(x)=\frac{x+1}{x^2-1}=\frac{1}{x-1}, x\neq \pm 1


So, such a function I simply came up with was

h(x)=x*tan(x+\frac{\pi}{2})

I used a graphing calculator to try understand what was happening around x=0, and it seems that

\underbrace{lim}_{x\rightarrow 0}h(x)=-1

Now I just want to understand why this limit tends to -1, not any other real values.

[arildno]

Well, you might try utilizing the identity:
tan(x+y)=\frac{\sin(x)\cos(y)+\cos(x)\sin(y)}{\cos (x)\cos(y)-\sin(x)\sin(y)}

[statdad]

A simpler example might be


f(x) = x \sin\left(\frac 1 x \right)


for which


\lim_{x \to 0^+} f(x) = 0

[Mentallic]

Aha

tan(x+y)=\frac{sin(x+y)}{cos(x+y)}

But all I get using this result is

tan(x+\frac{\pi}{2})=-cot(x)

It isn't helping just yet.

[Mentallic]

Hang on...

So the function now is f(x)=-\frac{x}{tan(x)}

and since the gradients of x and tanx at x=0 are equal, this gives it the value 1?

[arildno]

Indeed.

Or, as you can verify:
x\tan(x+\frac{\pi}{2})=-\frac{x}{\sin(x)}\cos(x)

[Mentallic]

Well, I remember the result

\lim_{x \to 0}\frac{x}{sin(x)}=1

and cos(0)=1 so I guess we can deduce that:

\lim_{x \to 0}-\frac{x}{sin(x)}cos(x)=-1

However, I'm sure that the function doesn't exist at the point x=0, so if I were to draw the function, I would leave an empty circle at the point (0,-1)?

Just like my previous mentioned function: f(x)=\frac{x+1}{x^2-1}
if I were to draw this function, I would quickly notice it is the same as f(x)=\frac{1}{x-1} except x\neq -1

Can I do the same for f(x)=-x cot(x) ? That is to say, can I find this equal to a simpler form (or more complicated if need be) of the same function, that instead is defined at x=0?