How would one go about solving this limit : $\lim_{x\to+0} ((x+9)^x-9^x)^x$ $\displaystyle \tag*{} \lim \limits _{x\to+0} ((x+9)^x-9^x)^x$
Hi everyone ! Sorry in advance for any formatting and grammatical errors !
My attempt to solve this limit went something like this :

*

*Write the limit as $\displaystyle \lim \limits_{x\to+0} e^{x \ln((x+9)^x-9^x)} $

*Move the limit to the power . Now I have to solve : $ \displaystyle \lim \limits _{x\to+0} x \ln((x+9)^x-9^x) $ .

*I realize this is $ 0 \cdot (-\infty)$ so I write it as : $\displaystyle \lim \limits _{x\to+0} \frac{\ln((x+9)^x-9^x)}{1/x} $.

*Now it's $-\frac{\infty}{\infty}$ and so I use L'Hôpital's rule.

*I get :  $\displaystyle \lim \limits _{x\to+0} \frac{((x+9)^x\left(\frac{x}{x+9}+\ln(x+9)\right)-9^x\ln9)(-x^2)}{(x+9)^x-9^x)} $

*Sad because this is $\frac{0}{0}$ and the only thing that I can do is apply L'Hôpital's rule again which would ( I think ) make things worse.

Wolfram Alpha gives 1 as the answer but no Step-by-Step solution sadly.
Any help appreciated !
 A: $\newcommand{\d}{\mathrm{d}}$As you say, we need to evaluate the logarithmic limit. Let's first try L'Hopital: $$\lim_{x\to0}\frac{\ln[(x+9)^x-9^x]}{1/x}=\lim_{x\to0}(-x^2)\cdot\frac{[\ln(x+9)+\frac{x}{x+9}](x+9)^x-\ln(9)\cdot 9^x}{(x+9)^x-9^x}$$No... this is horrible and very possibly many never resolve. But there is more to limits than L'Hopital (there is even a tag dedicated to this on the forum). Let's try: $$\ln[(x+9)^x-9^x]=\ln(9^x)+\ln\left[\left(\frac{x}{9}+1\right)^x-1\right]$$And let's recall that for $|x|\lt1$, $(1+x)^y=\sum_{n=0}^\infty\binom{y}{n}x^n$ Where appear the more generalised binomial coefficients. Notice that the first term is $1$ so the $-1$ cancels. I then can write the above as (for small $x$, but we want $x\to0$ anyway): $$x\cdot\ln(9)+\ln\left[\sum_{n=1}^\infty\binom{x}{n}\frac{x^n}{9^n}\right]$$Multiplying by $x$ and taking the limit, L'Hopital becomes useful again (I am casually splitting the limit across the addition and multiplication - this is because all individual limits exist, as I will show): $$\begin{align}\lim_{x\to0}x^2\cdot\ln(9)+\lim_{x\to0}x\cdot\ln\sum_{n=1}^\infty\binom{x}{n}\frac{x^n}{9^n}&=0+\lim_{x\to0}\frac{\ln(x^2/9+x^3(x-1)/162+\cdots)}{1/x}\\&\overset{\text{L.H.}}{=}\lim_{x\to0}(-x^2)\cdot\frac{2x/9+x^2(x-1)/54+\cdots}{x^2/9+x^3(x-1)/162+\cdots}\\&=\lim_{x\to0}\frac{x^3\cdot(-2/9)+O(x^4)}{x^2\cdot(1/9)+O(x^3)}\\&=0\end{align}$$Why zero at the end there? It is a standard limit fact that since the numerator polynomial dominates ($x^3\ll x^2$ as $x\to0$) the limiting behaviour is thus determined. Divide through by $x^2$ if you don't believe me!
Since the limit of the logarithm was $0$, the limit of the whole thing is $e^0=1$.
The $9$ was quite irrelevant here, by the way. Any positive constant would have done.
A: Let's start from understanding how $(x+9)^x-9^x$ behaves for $x\to 0$. You have
$$(x+9)^x-9^x = 9^x\left(\left(1+\frac{x}{9}\right)^x-1\right) = 9^x \left(e^{x\log(1+\frac{x}{9})}-1\right)\,.$$
Now $\log(1+\frac{x}{9}) = \frac{x}{9}+o(x)$ and so $$e^{x\log(1+\frac{x}{9})}= e^{\frac{x^2}{9}+o(x^2)} = 1 + \frac{x^2}{9}+o(x^2)\,.$$
On the other hand $9^x = 1+O(x)$ and so
$$(x+9)^x-9^x = \frac{x^2}{9}+o(x^2)\,.$$
Now we are left with studying the limit
$$\lim_{x\to 0}\left(\frac{x}{3}\right)^{2x}\,.$$
Clearly $3^{2x}\to 1$, so we just need to worry about $x^{2x} = e^{2x\log x}$. Now you have that $x\log x$ is a case of $0\times\infty$, as you said. However, from now it should be easier to use De L'Hôpital and conclude. Can you do it?
