<s>[INST] 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.
 [/INST] 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.
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