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what is the meaning of the formulas 1/0 and 0/0 by looking at them from the prospective of calculus and limits ? and as they are indeterminate forms , how is algebraic manipulation of a function that carries one of those formulas at some point capable of evaluating its limit ?

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  • $\begingroup$ If it is not determinate, it is indeterminate. $\endgroup$ – hamam_Abdallah Dec 3 '16 at 21:51
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You can get intuition by looking at

$$\lim_{x \to 0} ax/x$$

This is of course $a $ but simply taking limits in the numerator and denominator would give $0/0$. This example shows us that in these cases we need more information than just the limits of the numerator and denominator, we need to know how fast they go to these limits. Algebraic manipulation can isolate that information for us sometimes. (In this example, by just cancelling $x/x=1$ for nonzero $x $, though this can be more complicated sometimes).

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In calculus, such expressions are short-hand notations for limits. $1/0$ isn't indeterminate, but undefined. But in this context, it could be short-hand for a limit which diverges to $\pm \infty.$ $0/0$ is an indeterminate form, because in various situations, the limit it represents can be any real number or infinity.

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Consider the function $f(x)=\frac{\sin x}{x}$. A graph of $f(x)$ might look like:

Graph of f(x)

It seems that $f(0)=1$, but $f(0)=\frac{\sin 0}{0}=\frac{0}{0}$. Thus, $f(x)$ has an undefined point at $x=0$ - it is an $\frac{0}{0}$ indeterminate form!

Some indeterminate forms will not have a limit. For example,

$$\lim_{x\to 0}\frac{1}{x}$$

is undefined: it diverges to $\infty$ from the positive direction and $-\infty$ from the negative direction. But $\displaystyle\lim_{x\to 0}f(x)$ seems to be $1$. Indeed, this limit is shown here:

How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

Although this particular function required no algebraic manipulation, clearly more complicated functions would require it.

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