how to show that $ \ln (x^2 e^{-x(a-1)}+1) = O(x^2e^{-x(a-1)}) $

Question

I found this strange result from a solved exercise: (with $x\to\infty$ and $a>1$)

$$ \ln (x^2 e^{-x(a-1)}+1) = O(x^2e^{-x(a-1)}) $$

I cannot explain how to get this, I can only think of using Taylor series, I tried Maclaurin expansion of logarithm: $$ \ln (1+x) = x +o(x)$$ I can use this because if $x\to\infty$ and $a>1$ then $x^2 e^{-x(a-1)}\rightarrow 0 $, so I get:

$$ \ln (x^2 e^{-x(a-1)}+1) = x^2e^{-x(a-1)} + o(x^2e^{-x(a-1)}) $$

How can I arrive to $O(x^2e^{-x(a-1)})$ ? Any ideas?

Answer 1

Hint: $\displaystyle \lim_{t\to 0} \frac{t}{ln(t+1)}=1$.