Two different answers to $\lim_{x \to -\infty} \frac{8x^2-2x^3+1}{6x^2+13x+4}$

Question

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.

Answer 1

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.

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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.

Answer 2

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 $\infty$.

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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.