If I have $a=\dfrac{1-1}{1-1}$, is the solution $a=1$, since I can let $b=1-1$, and then I have $a=\dfrac{b}{b}=1$, or is it undefined since I have division by zero (or even else $a=0$ since the numerator is $0$ itself)?
If it makes any difference, I've arrived at this from: $$\lim_{x\rightarrow\pi}\dfrac{1+\cos^3x}{1-\cos^2x}$$
$\endgroup$ 22 Answers
$\begingroup$It is undefined because you're dividing by $0$.
Hint: As for the limit, note that $\forall z\in \mathbb R\left(z^3+1=(z+1)(z^2-z+1)\right)$.
$\endgroup$ $\begingroup$This is kind of like saying $a=\frac 00 = 1$ because if you let $b=0$, then $a=\frac bb = 1$. However, we know that $\frac 00 $ is undefined.
EDIT: Any number divided by zero is undefined.
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