The limit of the nth root of a to the n plus b to the n is the maximum of (a,b) I've been asked to prove the following from Spivak's Calculus
$$\lim_{n\to\infty}\sqrt[n]{a^n+b^n}=\max(a,b); a,b > 0$$
I understand that this is a proof by cases, and that our cases are $a=b$, $a>b$, and $b>a$. I have done the $a=b$ case, but I am stuck on the $a>b$ and $b>a$ cases.
Some hints would be appreciated-
Thanks!
 A: Hint: 
If $a \leqslant b$
$$b \leqslant (a^n + b^n)^{1/n} \leqslant 2^{1/n} b$$
A: we assume $a<b$.
Now use the sandwich to obtain: 
$b=\lim\limits_{n\to \infty} \sqrt[n]{b^n}\leq\sqrt[n]{a^n+b^n}\lim\limits_{n\to \infty} \sqrt[n]{a^n+b^n}\leq \lim\limits_{n\to \infty} \sqrt[n]{2b^n}=\lim\limits_{n\to \infty}\sqrt[n]{2}\lim\limits_{n\to \infty}\sqrt[n]{b^n}=1b$
A: HINT: Factor out the larger of the two numbers $a$ and $b$ (remembering that $\root n\of {a^n} = a$).
A: WLOG let $a>b$.
$$\begin{array}{rcl}
\displaystyle \lim_{n\to\infty}\sqrt[n]{a^n+b^n}
&=& \displaystyle \lim_{n\to\infty} a\sqrt[n]{1+\left(\dfrac b a\right)^n} \\
&=& \displaystyle \lim_{n\to\infty} a\left(1+\left(\dfrac b a\right)^n\right)^{1/n} \\
&=& \displaystyle \lim_{n\to\infty} a + \dfrac a n\left(\dfrac b a\right)^n - \dfrac{a(n-1)}{2n^2} + \cdots \\
&=& \displaystyle a + 0 + 0 + \cdots \\
&=& \displaystyle a \\
\end{array}$$
A: Suppose $a >b\ge 0$ then note that $\sqrt[n]{a^n+b^n} = a \sqrt[n]{1+({b \over a})^n}$.
Since $\sqrt[n]{1+({b \over a})^n} = e^{{1 \over n} \log (1+({b \over a})^n)}$, we see that $\sqrt[n]{1+({b \over a})^n} \to 1$ from which
the result follows.
