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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.
\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.
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