Evaluation of exponential limits

Question

Evaluate the limit $$\lim_{n \to ∞}(1+x^n)^\frac{1}{n}$$ For $x=0$ and $x=1$ the limit is $1$ Now for $0<x<1$ we have $x^n \to 0$ so the limit is $1$ Now for $x>1$ i dont know how to move forward.

Answer 1

For $x>1$ $$(1+x^n)^{\frac{1}{n}}=x\left(\frac{1}{x^n}+1\right)^{\frac{1}{n}}\rightarrow x.$$

Answer 2

In general, given $a_1,\dots,a_k$, we have $$\lim_{n \to \infty} \sqrt[n]{a_1^n+\cdots+a_k^n} = \max(a_1,\dots,a_k)$$by squeezing principle.

Answer 3

If you expand this using the Binomial series to two terms you get

$$ \lim_{n\rightarrow\infty} 1 + \frac{x^n}{n}$$

The second term evaluates to an indeterminate form, so using L'Hopital's rule you get

$$ \lim_{n\rightarrow\infty} 1 + \frac{x^n\ln{x}}{1}$$

This shows you that if $|x| < 1$ then the limit will evaluate to 1 as the second term goes to zero for the numerator being both positive/negative and will otherwise go to $-\infty/\infty$ if $x > |1|$ in an oscillatory way unless n is restricted to even/odd values.

Answer 4

Let $x>1, n \ge 1; $

$x:=1+y$, $y >0;$

$ x^n =(1+y)^n \gt 1+ ny >1$.

$1 \lt (1+x^n)^{1/n} \lt (2x^n)^{1/n} =2^{1/n}x ;$

Take the limit.