Two different answers to $\lim_{x \to -\infty} \frac{8x^2-2x^3+1}{6x^2+13x+4}$ Let us evaluate
$$\lim_{x \to -\infty} \frac{8x^2-2x^3+1}{6x^2+13x+4}$$
Dividing the numerator and denominator by $x^3$ we will be left with
$$\lim_{x \to -\infty} \frac{\frac{8}{x}-2+\frac{1}{x^3}}{\frac{6}{x}+\frac{13}{x^2}+\frac{4}{x^3}}$$ and since all the terms containg $x$ term will go to $0$,our resultant limit becomes $-\frac{2}{0}=-\infty$.
But if we do it in another way substituting $x=-t$,then our limit becomes after dividing both numerator and denominator by $t$,
$$\lim_{t \to \infty} \frac{\frac{8}{t}+2+\frac{1}{t^3}}{\frac{6}{t}-\frac{13}{t^2}+\frac{4}{t^3}}$$
Here also since terms containing $t$ go to $0$,we are left with $\frac{2}{0}=+\infty$.
Why are we getting two different answers? Surely one of the method is invalid. In books,the first answer was marked correct. But i want to know what's wrong with the second approach.
 A: The problem is that you cannot determine the sign of the limit $\frac 20$ because in this simplified form you have no information on "the sign of $0$"
i.e. you lost the information about how the denominator goes to zero.
Does the denominator goes to $0^+$ or $0^-$ or does it goes to zero oscillating around ?
In this kind of problem it is better to factor the dominant term :
$\dfrac{8x^2-2x^3+1}{6x^2+13x+4}=\underbrace{\dfrac{-2x^3}{6x^2}}_{\to+\infty}\times\underbrace{\dfrac{1+\frac 4x-\frac 1{2x^3}}{1+\frac{13}{6x}+\frac 2{3x^2}}}_{\to 1}\to+\infty$
Why do we do that ?
Because you'll learn later equivalents, i.e. $f(x)\sim g(x)$ if their ratio has limit $1$ and we can multiply and divide equivalents like this:

*

*numerator $8x^2-2x^3+1\sim -2x^3$

*denominator $6x^2+13x+4\sim 6x^2$
Basically you ignore all negligible terms and keep only the dominant term.
And you will write directly $\dfrac{8x^2-2x^3+1}{6x^2+13x+4}\sim\dfrac{-2x^3}{6x^2}\sim -\dfrac{x}{3}\to+\infty$
If you proceed by factorizing the dominant term now, equivalents will appear as a natural extension when you'll learn them formally later.

In fact apply the method to the fraction obtained after division by $x^3$ you get in the first case:

*

*dominant term on numerator $-2$

*dominant term on denominator $\frac 6x<0$
So the $-\frac20$ is actually $\dfrac {-2}{0^-}\to+\infty$ by the rule of signs.
While in the second method it is:

*

*dominant term on numerator $2$

*dominant term on denominator $\frac 6t>0$
So the $\frac20$ is actually $\dfrac {2}{0^+}\to+\infty$ as well.
And results are consistent.
A: Note that

*

*for $x\to -\infty$
$$\frac{6}{x}+\frac{13}{x^2}+\frac{4}{x^3} = \frac{1}{x}\left(6+\frac{13}{x}+\frac{4}{x^2}\right)\to 0 \cdot6=0$$
with $\frac 1x <0$, then $0$ is reached from the left side ($0^-$).

*

*for $t=-x\to \infty$
$$\frac{6}{t}-\frac{13}{t^2}+\frac{4}{t^3} = \frac{1}{t}\left(6-\frac{13}{t}+\frac{4}{t^2}\right)\to 0 \cdot6=0$$
with $\frac 1t >0$, then $0$ is reached from the right side ($0^+$).
Therefore in both cases limit is $\infty$.

Note that the second approach is fine indeed from here
$$\lim_{x \to -\infty} \frac{8x^2-2x^3+1}{6x^2+13x+4}=\lim_{t \to \infty} \frac{8t^2+2t^3+1}{6t^2-13t+4}$$
we can proceed as follows
$$\frac{8t^2+2t^3+1}{6t^2-13t+4}=\frac{t^3}{t^2}\cdot \frac{\frac8t+2+\frac1{t^3}}{6-\frac{13}t+\frac 4{t^2}}=t\cdot \frac{\frac8t+2+\frac1{t^3}}{6-\frac{13}t+\frac 4{t^2}}$$
from wich we obtain $\infty$ as limit.
Note also that in a similar way for the first one
$$\frac{8x^2-2x^3+1}{6x^2+13x+4}=x\cdot \frac{\frac 8 x-2+\frac 1{x^3}}{6+\frac {13} x+\frac 4{x^2}}$$
wich also leads to $\infty$ as limit.
