Infinite false limit verification via $N-M$ method

$$\lim_{x\to -\infty} \frac{x^3 + 6}{2x^2-5} = +\infty$$

The limit is false. Let's verify it.

By the formal definition, $\forall N > 0, \exists M_N > 0$ such that $\forall x < -M_N$ we have $f(x) > N$. Whence

$$\frac{x^3+6}{2x^2-5} > N$$

To deal with this, I thought about this

$$x^3 + 6 \leq x^3/3$$ $$2x^2 - 5 \leq 2x^2$$

remembering $x \to -\infty$. Hence

$$x > \underbrace{6N}_{M_N}$$

which means using the definition: $$x < -6N$$

But this seems legit, as for $M$ to grow big, $-6N$ grows infinitely negatively big...

Where am I wrong?

Solution 1:

The error in your approach lies in the fact that from the inequalities$$x^3+6\leqslant\dfrac{x^3}3\tag1$$and$$2x^2-5\leqslant2x^2$$there is nothing that you can deduce about the quotient $\dfrac{x^3+6}{2x^2-5}$. However, from $(1)$ and $2x^2-5\geqslant x^2$ (which holds when $x\notin\left[-\sqrt5,\sqrt5\right]$), you can deduce that$$\frac{x^3+6}{2x^2-5}\leqslant\frac x3.\tag2$$So, and since $\lim_{x\to-\infty}\dfrac x3=-\infty$, $\lim_{x\to-\infty}\dfrac{x^3+6}{2x^2-5}=-\infty$ too. On the other hand, it follows from $(2)$ that$$x<-\sqrt5\implies\dfrac{x^3+6}{2x^2-5}<0,$$and so you cannot possibly have$$\lim_{x\to-\infty}\dfrac{x^3+6}{2x^2-5}=\infty.$$