Indeterminant form in a limit

Question

If, after evaluating a limit we get an indeterminant, of the form $\dfrac{-1}{0}$, how do we conclude whether it converges to $+\infty$ or $-\infty$? Just because it is negative, how can we conclude for sure that it will tend to $-\infty$?

Please refer to the following picture for the original question

After evaluating this I have an indeterminant of the form $\dfrac{-1}{0}$

Answer 1

We can't. Take, for instance, the limits$$\lim_{x\to0^+}\frac{x-1}{-x}\text{ and }\lim_{x\to0^+}\frac{x-1}x.$$In both cases, we have a limit of the type that you mentioned. But the first limit is $+\infty$, whereas the second one is $-\infty$.

Answer 2

write your term in the form $$\frac{x^3(-1+\frac{8}{x^3})}{x^2(2+\frac{5}{x}+\frac{7}{x^2})}$$ from here we get $$\lim_{x\to \infty}x\cdot \lim_{x\to\infty}\frac{-1+\frac{8}{x^2}}{2+\frac{5}{x}+\frac{7}{x^2}}$$ and you will get $$\infty\cdot (-\frac{1}{2})=-\infty$$

Answer 3

It is case by case:

suppose we have $f(x)=\frac{-1}{x^2}$,in that case we know that $\lim_{x\to 0}f(x)\to -\infty$, Because $\forall x\in\Bbb R ,\,x^2>0$.

But for example $f(x)=\frac{-1}{x}$ we can't say anything, because the left side limit(the limit as $x$ goes to $0$ from the "negative"(left side) side) is $\infty$ and the right side limit is $-\infty$.