Question

Calculate the following limit:


`\displaystyle \lim_(x \to \infty){(\log(x^8 - 5))/(x^2)}`

Answer 1

If we evaluate at infinity, we have:

`\bf{\displaystyle L = \lim_(x \to \infty){(\log(x^8 - 5))/(x^2)} = (\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) = (\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) = (\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 (\log(x^8 - 5))/(x^2) \leq (\log(x^8))/(x^2) }`

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

`\boldsymbol{\displaystyle (\log(x^8))/(x^2) = (\log(e^(8y)))/(e^(2y)) = (8y)/(e^(2y))}`

A very important property about the exponential function is

`\boldsymbol{\displaystyle e^x > (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) > ((2y)^2)/(2!) = (4y^2)/(2) = 2y^2}`

From this it follows that

`\boldsymbol{\displaystyle (1)/(e^(2y)) < (1)/(2y^2) }`

Therefore, we have to

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