Asymptotic expression of $(1+bx^a)^n$ as $x\to0$

Question

Using Taylor expansion, we have $(1+x)^n\to 1+nx$ as $x\to0$. Now if I have a more general expression, say, $(1+bx^a)^n$, then does this expression still have such asymptotic behavior?

Any help is appreciated.

Answer 1

Yes. If you know that $(1 + Y)^n \sim 1 + nY$ as $Y \to 0$, and you call $Y = b x^a$ (for $a > 0$), it remains true that $x \to 0 \implies Y \to 0$.

The full expansion is the binomial expansion, so you really know that $$ (1 + Y)^n = \sum_{k \geq 0} {n \choose k} Y^k.$$ The first two terms are the terms you give. The remaining terms (finitely many if $n$ is an integer) all have larger powers of $Y$, and as $Y \to 0$ they become very small very quickly.