Simple Calculus Proof [duplicate]

Solution 1:

Suppose that $b \ge a$. Then $$c_n = b ((a/b)^n + 1)^{1/n}.$$ Since $$1 \le ((a/b)^n + 1)^{1/n} \le 2^{1/n}$$ and $2^{1/n} \to 1$ as $n \to \infty$ you can use the squeeze theorem to obtain the limit.

Solution 2:

Hint: factor out the biggest of the two; you will get $\max(a,b)\cdot x_n$, where $x_n\to 1$.

• If $a=b$, then $$ (a^n+b^n)^{1/n} = (2a^n)^{1/n} = 2^{1/n} a \xrightarrow[n\to\infty]{} 1\cdot a = a. $$

• If $a\neq b$, suppose (without loss of generality) that $a > b$.

$$ (a^n+b^n)^{1/n} = \left(a^n\left(1+\frac{b^n}{a^n}\right)\right)^{1/n} = a\left(1+\frac{b^n}{a^n}\right)^{1/n} \xrightarrow[n\to\infty]{} a\cdot 1 = a. $$ (using the fact that $0<\frac{b}{a}< 1$)

Solution 3:

There is a much shorter method than the one below. Let me present the idea. If $b>a$:

$$\sqrt[n]{a^n+b^n}=b\sqrt[n]{\left(\frac ab\right)^n+1} \Rightarrow \lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}=b \lim_{n \rightarrow \infty}\sqrt[n]{\left(\frac ab\right)^n+1}$$ $$\Rightarrow \lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}= b\sqrt[n]{\left(\lim_{n \rightarrow \infty}\frac {a^n}{b^n}\right)+1}=b$$

If $a>b$:

$$\sqrt[n]{a^n+b^n}=a\sqrt[n]{\left(\frac ba\right)^n+1} \Rightarrow \lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}=a \lim_{n \rightarrow \infty}\sqrt[n]{\left(\frac ba\right)^n+1}$$ $$\Rightarrow \lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}= a\sqrt[n]{\left(\lim_{n \rightarrow \infty}\frac {b^n}{a^n}\right)+1}=a$$ If $a=b$: $$\lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}=a\lim_{n \rightarrow \infty}2^{1/n}=a2^0=a=b$$ In either case we can see that the maximum of $a,b$ happens to be the limit. Thus: $$\lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}= \max (a,b)\sqrt[n]{\left(\lim_{n \rightarrow \infty}\frac {(\min(a,b))^n}{(\max(a,b))^n}\right)+1}=\max (a,b)$$