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$\lim_{x\to 0} \frac{x}{x}$ is an indeterminate form whereas $\lim_{x\to 0} \frac{[x^2]}{x^2}$ is not an interminate form (where $[x]$ represents the greatest integer function

Why is $\lim_{x\to 0} \frac{[x^2]}{x^2}$ not in indeterminate form?

As far as I know, $\frac{[x^2]}{x^2}$ gives a $\frac{0}{0}$ at $0$. What is the exact definition of 'indeterminate form'? Wikipedia does not help answer this question.

Yashas
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    How do they define "indeterminate form"? At first blush, they both seem indeterminate to me, since at $x=0$, they both are $0/0$. For $ x \ne 0$, the first expression reduces to 1, the second reduces to 0, but I see no fundamental difference between them. Where did you find that quote? – NickD Mar 23 '17 at 14:43
  • Exactly, that's how I see it too. I found it in my differential calculus textbook. – Yashas Mar 23 '17 at 14:57

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Indeterminate forms of functions are those forms which really have a finite value but it cannot be determined by just looking at it, for example $\frac{x}{x}$ at $x=1$ has value $1$, clearly. But $\lim _{x \rightarrow 0} \frac {x}{x}$ can be found using some algorithm i.e. generally cancelling of indeterminate factors. Indeterminate forms are of forms such as $$ \frac{(\rightarrow 0)}{(\rightarrow 0)} , \frac{(\rightarrow \infty)}{(\rightarrow \infty)}, {\rightarrow 1}^{ \rightarrow \infty }, {\rightarrow 0}^{ \rightarrow 0}$$

In your second example it is $\lfloor x^2 \rfloor$ , which at $x \rightarrow 0$ is exactly $0$. Therefore : $$\lim_{x\to 0} \frac{\lfloor x^2 \rfloor}{x^2}=\frac{0}{(\rightarrow 0)} = 0$$

On the other hand undefined form (although you haven't asked for this, I am telling ) are those which do not have really a finite value those aren't even defined to have finite value , for example : $$\frac {0}{0} ,{0}^{0 } , \frac{\infty}{\infty}$$

Hope it clears your doubt.

Jaideep Khare
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Note that in general $\displaystyle{\lim_{x\to a}}f(x)$ is independent of the value of $f(a)$ unless $f$ is continuous at $x=a$.

In the case of $f(x)=\dfrac{\lfloor x^2\rfloor}{x^2}$, $f(x)=0$ on the interval $(-1,0)$ and on the interval $(0,1)$, so $\displaystyle{\lim_{x\to0}}f(x)=0$ even though $f(0)$ is undefined.

This is, however, an indeterminate form since $\displaystyle{\lim_{x\to0}}\lfloor x^2\rfloor=0$ and $\displaystyle{\lim_{x\to0}} x^2=0$. Accordingly one may apply L'Hopital's rule:

\begin{eqnarray}\lim_{x\to0}\frac{\lfloor x^2\rfloor}{ x^2} &=& \lim_{x\to0}\frac{0}{2x}\\ &=&0 \end{eqnarray}

Here is a graph of the function:

enter image description here

John Wayland Bales
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I agree that

$$\lim_{x \to 0} \frac{\lfloor x^2 \rfloor}{x^2} $$

is clearly of indeterminate form. Even

$$\lim_{x \to 0} \frac{0}{x^2} $$

is of indeterminate form.

Without context, I imagine what happened is that the tendency to replace $\frac{0}{x^2}$ with $0$ is so strong that your source did so mentally before evaluating whether or not the limit was of indeterminate form.