Question

Calculate the following limit:
$\displaystyle \lim_{x \to \infty}{\dfrac{\log(x^8 - 5)}{x^2}}$

Answer 1

If we evaluate at infinity, we have:

                        $\bf{\displaystyle L = \lim_{x \to \infty}{\frac{\log(x^8 - 5)}{x^2}} = \frac{\infty}{\infty} }$

However, the infinity of the denominator has a higher order. Therefore, we can conclude that  $\boldsymbol{L = 0.}$

However, proving that the limit is 0 without using L'Hopital or the "order" criterion is complicated. To do so, let us denote:

                           $\boldsymbol{\displaystyle f(x) = \frac{\log(x^8 - 5)}{x^2} }$

To find the limit, we must look for two functions h(x) and g(x) such that h(x)≤ f(x)≤ g(x) and

                               $\boldsymbol{\displaystyle \lim_{x \to \infty}{h(x)} = 0, \qquad \lim_{x \to \infty}{g(x)} = 0}$

If we find these functions, then we can conclude that $\bf{\lim_{x \to \infty}{f(x)} = 0.}$

First, let's note that when x⁸ - 5 > 1, then log(x⁸ - 5) > 0 (and this is true when x is large). Likewise, we have that x² > 0 for x > 0. Therefore, we have:

                                   $\boldsymbol{\displaystyle f(x) = \frac{\log(x^8 - 5)}{x^2} \geq 0}$

when x "is big enough". Thus, we have h(x) = 0 where it is clear that $\bf{\lim_{x \to \infty}{h(x)} = 0.}$

To find the second function, let's first note that \log is an increasing function, so since x⁸ ≥ x⁸ - 5, then log(x⁸) ≥ log(x⁸ - 5). So we have to

                $\boldsymbol{\displaystyle \frac{\log(x^8 - 5)}{x^2} \leq \frac{\log(x^8)}{x^2} }$

now, if we take y = e^y, then we can write

                  $\boldsymbol{\displaystyle \frac{\log(x^8)}{x^2} = \frac{\log(e^{8y})}{e^{2y}} = \frac{8y}{e^{2y}}}$

A very important property about the exponential function is

                    $\boldsymbol{\displaystyle e^x > \frac{x^n}{n!}}$

For any n $\bf{n \in \mathbb{N}}$ and x > 0. If we take n = 2, then we have

                    $\boldsymbol{\displaystyle e^{2y} > \frac{(2y)^2}{2!} = \frac{4y^2}{2} = 2y^2}$

From this it follows that

                     $\boldsymbol{\displaystyle \frac{1}{e^{2y}} < \frac{1}{2y^2} }$

Therefore, we have to

                    $\boldsymbol{\displaystyle \frac{\log(x^8 - 5)}{x^2} \leq \frac{\log(x^8)}{x^2} < \frac{8y}{2y^2} = \frac{4}{y} = \frac{4}{\log x} }$

yes, $\bf{g(x) = 4/\log x}$ where $\bf{\lim_{x \to \infty}{g(x)} = 0}$. Also, $\bf{h(x) \leq f(x) < g(x)}$. Therefore, $\bf{\lim_{x \to \infty}{f(x)} = 0}$.

$\red{\boxed{\green{\boxed{\boldsymbol{\sf{\purple{Pisces04}}}}}}}$