L'Hospital's rule for $\lim_{n\rightarrow \infty} (a^n + b^n)^\frac{1}{n}$. I have recently come across a problem concerning the limit of a sequence given as, $$\lim_{n\rightarrow \infty} (a^n + b^n)^\frac{1}{n}$$
Where $0<a<b$. Solving the limit is not a concern. But applying L'Hospital's rule to this limit is one to me. 
On rearranging we get, $b \lim_{n\rightarrow \infty} ((\frac{a}{b}) ^n + 1)^\frac{1}{n}$. The solver had then applied the L'Hospital's rule to evaluate the limit. Here, as $n\rightarrow \infty$, the limit approaches the form $1^0$. I have tried to think about this limit approaching an indeterminate form. It does not. From my knowledge of Cauchy's Mean Value theorem, I say that we cannot apply the rule to determinate forms. But it seems to me that we can because it worked for the solver. I think I am not clear on this. Can somebody help? 
 A: Why not $$y= (a^n + b^n)^\frac{1}{n}\implies \log(y)=\frac{\log(a^n + b^n)}{n}$$ and then $$\frac{a^n \log (a)+b^n \log (b)}{a^n+b^n}$$ Now, divide everything by $b^n$ since $b>a$ to get the limit of $\log(y)$ and then the limit of $y$.
A: Note that $\log(1+x)\le x$ for $x>-1$.  Therefore we have for $a<b$
$$1\le \left(1+\left(\frac ab\right)^n\right)^{1/n}\le e^{\frac1n \left(\frac ab\right)^n}\tag 1$$
whence applying the squeeze theorem to $(1)$ yields the coveted limit
$$\lim_{n\to \infty}\left(1+\left(\frac ab\right)^n\right)^{1/n}=1$$
And we are done!
A: One can take a different view by an expansion which is
$$(1 + x)^{1/n} = 1 + \frac{1}{n} \, \frac{x}{1!} + \frac{1}{n} \, \left( \frac{1}{n} - 1\right) \, \frac{x^{2}}{2!} + \mathcal{O}\left(\frac{1}{n^{3}}\right)$$
for which, $0 < a <b$,
$$\left(1 + \left(\frac{a}{b}\right)^{n} \right)^{1/n} = 1 + \frac{1}{n} \, \left(\frac{a}{b}\right)^{n} + \frac{1}{2 \, n} \, \left( \frac{1}{n} - 1\right) \, \left(\frac{a}{b}\right)^{2n} + \mathcal{O}\left(\frac{1}{n^{3}}\right)$$
and
\begin{align}
\lim_{n \to \infty} (a^n + b^n)^{1/n} &= b \, \lim_{n \to \infty} \left(1 + \left(\frac{a}{b}\right)^{n} \right)^{1/n} = 1.
\end{align}
If $a=b$ then
$$ 2^{1/n} = 1 + \frac{1}{n} + \frac{1}{2 \, n} \, \left( \frac{1}{n} - 1\right)  + \mathcal{O}\left(\frac{1}{n^{3}}\right)$$
which leads to
$$\lim_{n \to \infty} 2^{1/n} = 1.$$
