I was reading about limits and stumbled upon this site and saw the author doing something funny:
$$ \lim_{x\rightarrow 0} \frac{x^3-7x}{x^3} $$
It is quite clear that you can apply l'Hospital at this point but the author does something else:
$$ \lim_{x\rightarrow 0} \frac{x\left(x^2-7\right)}{x\left(x^2\right)} $$
$$ \lim_{x\rightarrow 0} \frac{x^2-7x}{x^3} = \quad'' \quad \frac{-7}{0} \quad'' = -\infty $$
I looked up indeterminate forms and could not find anything like this. What is going on here ?
With l'Hospital I dont get much farther either.
Answer
L'Hospital's rule is unnecessary here, as simple algebra gets rid of the indeterminate form. Just cancel an $x$ from the top and bottom as the author does:
$$
\lim_{x\rightarrow 0} \frac{x^3 - 7x}{x^3} = \lim_{x\rightarrow 0} \frac{x^2 - 7}{x^2}.
$$
Now, when the author writes "$-7/0^+$", he's saying that the limit of the numerator is -7, and the limit of the denominator is 0, but is positive on both sides. Whenever you have a finite limit for the numerator and a 0 limit for the denominator, the limit will either be nonexistent (if the signs don't match on opposite sides) or $\pm \infty$ (if they do). In this case the denominator is positive on both sides, so, since the numerator is negative, the limit will be $-\infty$. It is this sense in which $-7/0^+ = -\infty$. Similarly, in this sense $-7/0^- = \infty$, while $-7/0$ remains undefined.
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