Understanding Undefined Values at x=a in a Function

[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$\neq$0.

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?