graphing $y=\lim\limits_{n \to \infty}{\sqrt[n]{1+x^n}}$

Question

I'm trying to graph the following equation

$$y=\lim\limits_{n \to \infty}{\sqrt[n]{1+x^n}}$$ ($x \geq 0$)

$x = 0$: $y = \lim\limits_{n \to \infty}{\sqrt[n]{1+0^n}} = \lim\limits_{n \to \infty}{\sqrt[n]{1}} = 1$

$0 < x < 1$: $y = \lim\limits_{n \to \infty}{\sqrt[n]{1+\frac{1}{\alpha}^n}} = \lim\limits_{n \to \infty}{\sqrt[n]{1}} = 1$

$1 \leq x$: $y = \lim\limits_{n \to \infty}{\sqrt[n]{\alpha}} = 1$

It seems like $y = 1$ all the time.

However, shouldn't there also be an asymptote $y=x$? As $x \to \infty$: $y = \lim\limits_{n \to \infty}{\sqrt[n]{x^n}} = x$, right?

If I am to come to the conclusion that the graph is equivalent to $y=1$ and that the asymptote is to be ignored, I must prove that $\lim\limits_{n \to \infty}{\sqrt[n]{1+x^n}} = 1$. Is this equality true for $x \geq 0$?

Answer 1

Hint:

If $x\ge1$, then

$$x\sqrt[n]2=\sqrt[n]{x^n+x^n}\ge\sqrt[n]{1+x^n}>\sqrt[n]{x^n}=x$$

Now apply squeeze theorem.

If $x<1$, then

$$\sqrt[n]{1+x^n}=x\sqrt[n]{1+\frac1{x^n}}=x\sqrt[n]{1+y^n}$$

where $y=\frac1x>1$. Now apply the first case.

Answer 2

For $x \geq 0$. What we have is,

$$\sqrt[n]{1+x^n}$$

$$=\sqrt[n]{x^n(1+\frac{1}{x^n})}$$

$$=x\sqrt[n]{1+\frac{1}{x^n}}$$

By the Binomial series,

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

Suppose $x \geq 1$. It is clear that as $n \to \infty$ that this tends to

$$x$$ Suppose that $x \in (0,1)$. Then $x=\frac{1}{y}$ for some $y>1$. Then our expression becomes equivalently,

$$x\sqrt[n]{1+y^n}$$

We have just seen that for $y>1$ we have,

$$\sqrt[n]{1+y^n} \to y$$

So then the limit tends to,

$$xy=1$$

The case $x=0$ is clear.

So we have

$$y=\begin{cases} 1 & x \in [0,1) \\ x & x \in [1,\infty) \end{cases}$$

Answer 3

Using Mathematica:

Plot[Limit[(1 + x^n)^(1/n), n -> Infinity], {x, -3, 3}]