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How do I show that $$lim_{ntoinfty} sqrt[n]{a^n+b^n}=operatorname{max}(a,b)$$
with $a,bge0$.

I tried to do this by dividing it in two cases, when $a=b$ and $agt b$.

In the case $agt b$ I factored $a^n$ like this: $$lim_{ntoinfty} sqrt[n]{a^n left(1+{b^nover a^n} right)} = lim_{ntoinfty} asqrt[n]{1+{b^nover a^n}} = alim_{ntoinfty} sqrt[n]{1+{b^nover a^n}}$$

Then I expressed it in exponential way. $$alim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}$$

Now I need to prove that $lim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}=1$ so the whole limit is equal to a. The problem is that I don't know how to take the limit of $lim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}$ I tried to use the natural logarithm, but it ended up like this:
$$lim_{ntoinfty} logleft({1+{b^nover a^n}}right)^{1/n}=lim_{ntoinfty} {1over n} logleft({1+{b^nover a^n}}right)=lim_{ntoinfty} {1over n} lim_{ntoinfty}logleft({1+{b^nover a^n}}right)=0$$
What did I do wrong and how can I do it right?

Assume that $a geq b$. Else, relabel the numbers $a leftrightarrow b$. This being understood, we have $a = max(a,b)$. Then $$a =sqrt[n]{a^n} leq sqrt{a^n+b^n} leq sqrt{a^n + a^n} = sqrt[n]{2} a. $$Now apply $lim_{n to +infty}$ on everything, noting that $sqrt[n]{2} to 1$. It follows from the squeeze theorem that $$lim_{nto +infty}sqrt[n]{a^n+b^n} = a,$$ as wanted.

For your specific question you may proceed as follows using

• 
  $lim_{xto 0}left(1+x right)^{frac{1}{x}} = e$ and


• Let $a>b > 0$ (For $b= 0$ there is nothing to show.) $Rightarrow 0< q:= frac{b}{a} < 1 Rightarrow q^n stackrel{n to infty}{longrightarrow} 0$

begin{eqnarray*} left({1+{b^nover a^n}}right)^{1/n}
& = & (1+q^n)^{frac{1}{n}}\
& = & left( underbrace{(1+q^n)^{frac{1}{q^n}}}_{stackrel{n to infty}{longrightarrow}e} right)^{underbrace{frac{q^n}{n}}_{stackrel{n to infty}{longrightarrow}0}}\
& stackrel{n to infty}{longrightarrow} & e^0 = 1
end{eqnarray*}