Removing the subtractive cancellation of $(a+x)^n - a^n$ as x approaches zero

Question

$$(a+x)^n - a^n = \\ e^{n\ln(a+x)}-e^{n\ln(a)} = \\ ?$$

This looks simple but I'm stumped.

I've tried to solve this another way too:

$$ f(a) = a^n \\ \frac{\delta f}{\delta a} = \lim_{x\rightarrow 0}\frac{(a+x)^n-a^n}{x} \\ \\ (a+x)^n - a^n = \frac{(a+x)^n-a^n}{x}x = \frac{\delta f}{\delta n}x=a^n\ln(a)x:x\rightarrow0\\$$

How do I finish the first one, and is the second one correct?

Answer 1

Hint: Apply the binomial theorem: $$(a+x)^n=a^n+nxa^{n-1}+\cdots+nax^{n-1}+x^n.$$ Your second method is incorrect because if you take $f(x)=x^n$, then $f'(x)=nx^{n-1}$. You are taking $f(x)=a^x$.

Answer 2

$$S_n=(a+x)^n-a^n=\sum_{k=1}^n \binom{n}{k} a^{n-k} x^k=a^n\sum_{k=1}^n \binom{n}{k}\left(\frac{x}{a}\right)^k$$ Truncate where ever you want.

If $x \ll a$, the first approximation would be $$S_n=n a^{n-1} x$$ but you can have better with $$S_n=\frac{2 n x a^n}{2 a-(n-1)x}$$