Proving that a function $f(x) = \frac{x^2-1}{|x^3-1|} $ is bounded and find its upper and lower bound

Question

How do I prove that a function:

$$ f(x) = \frac{x^2-1}{|x^3-1|}\\ x \in \mathbb{R}\\ x \ne 1 $$

is bounded and find its upper and lower bounds?

What I've tried:

This can be proven graphically and from the graph it's obvious that the function is indeed bounded and the bounds are $\frac{2}{3}$ and $-1$. One can also see that $|x^3-1| \geq |x^2-1|$ for every value of $x$, so $f(x) \leq 1$. But the graph shows that the value of $f(x)$ is less than $\frac{2}{3}$.

Another try was to assume that there exist values $m$ and $M$ such that $m \leq f(x) \leq M$ and play with the inequalities, but eventually i came to nothing.

How do I analytically find upper and lower bounds of a function (and in particular those for the function above) without using derivatives?

Answer 1

On boundedness.

For $0\ne x\in \Bbb R.$ We have $|f(x)|=|\frac {x+1}{x^2+x+1}|=\frac {|x+1|}{|x^2+x+1|}.\quad$ We have

$(\bullet)... $ $x^2+x+1=(x+\frac {1}{2})^2+\frac {3}{4}\geq \frac {3}{4}.$ So for all $x$ we have $|x^2+x+1|=x^2+x+1\geq \frac {3}{4}.$

Version 1:

(I). For $|x+1|\leq 2$ we have $|f(x)|\leq \frac {2}{|x^2+x+1|}=$ $\frac {2}{x^2+x+1}\leq$ $ \frac {2}{3/4}=$ $\frac {8}{3}.$

(II). For $|x+1|>2$ we have $\frac {1}{|f(x)|}=$ $|\frac {x^2+x+1}{x+1}|=$ $|\frac {(x+1)^2-(x+1)+1}{x+1}|=$ $|(x+1)-1+\frac {1}{1+x}|\geq$ $ |(x+1)|-|(-1)|-|\frac {1}{1+x}|\geq$ $ 2-1-\frac {1}{2}=$ $\frac {1}{2},$ implying $ |f(x)|\leq 2.$

(III). Therefore by (I) and (II) we have $|f(x)\leq \max (\frac {8}{3}, 2)=\frac {8}{3}$ for all $x.$

Version 2: From a general perspective.

We have $\lim_{|x|\to \infty}|f(x)|=0 .$ So for some $K>1$ we have $|x+1|>K \implies |f(x)|<1.$ Therefore:

(I'). $|x|>K+1\implies$ $ |x+1|>K\implies |f(x)|<1.$

(II'). By $(\bullet) $ we have $|x|\leq K+1\implies$ $ |f(x)|\leq \frac {|x+1|}{3/4}\leq$ $ \frac {|x|+1}{3/4}\leq$ $ \frac {K+2}{3/4}=$ $\frac {4}{3}(K+2).$

(III'). So for all $x$, by (I') and (II') we have $|f(x)|\leq \max (1, \frac {4}{3}{K+2})=$ $\frac {4}{3}(K+2).$

These upper bounds for $|f|$ may not be very "sharp", but when trying to stay below $\infty ,$ any finite value will do.

Answer 2

The numerator and the denominator are continuous functions. Over any finite interval each is bounded.

When the denominator equals 0 (at x = 1) the $\lim_\limits{x\to 1} \frac {f(x)}{g(x)}$ exists and its bounded.

The rational function is bounded over any finite interval.

And an the tails we have a higher degree polynomial in the denominator.

$\lim_\limits{x\to \infty} \frac {f(x)}{g(x)} = 0$ and $\lim_\limits{x\to -\infty} \frac {f(x)}{g(x)} = 0$