If we evaluate at infinity, we have:
However, the infinity of the denominator has a higher order. Therefore, we can conclude that
However, proving that the limit is 0 without using L'Hopital or the "order" criterion is complicated. To do so, let us denote:
To find the limit, we must look for two functions h(x) and g(x) such that h(x)≤ f(x)≤ g(x) and
If we find these functions, then we can conclude that
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:
when x "is big enough". Thus, we have h(x) = 0 where it is clear that
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
now, if we take y = e^y, then we can write
A very important property about the exponential function is
For any n
and x > 0. If we take n = 2, then we have
From this it follows that
Therefore, we have to
yes,
where
. Also,
. Therefore,
.