What is the limit of $\lim_{x \to \infty}\sqrt[x]{2^x+3^x+4^x}$

Question

I would like to ask for some help regarding the limit below

$$\lim_{x \to \infty}\sqrt[x]{2^x+3^x+4^x}$$

Am i supposed to use the Squeeze theorem?

Answer 1

Note that we have for a vector $\vec{x}\in\mathbb{R}^{n}$:

$$\|\vec{x}\|_{p}=\sqrt[p]{\sum_{k=1}^{n}x_{k}^{p}}$$

And that we have the supremum norm:

$$\|\vec{x}\|_{\infty}=\lim_{p\to\infty}\|\vec{x}\|_{p}=\max\{|x_{1}|,\cdots,|x_{k}|\}$$

Therefore in your case $\vec{x}\in\mathbb{R}^{3}$, with $\vec{x}=\left(\begin{smallmatrix}2 \\ 3 \\ 4\end{smallmatrix}\right)$, we have:

$$\lim_{n\to\infty}\left(\sqrt[n]{2^{n}+3^{n}+4^{n}}\right)=\|\vec{x}\|_{\infty}=\max\{2,3,4\}=4$$

Answer 2

Hint: Factor $4^x$ inside the radical.

Answer 3

The dominating term in the radical is $4^x$. So, we have $L=4^{(x/x)} = 4$

Answer 4

In my mind, Shaktal's solution is the most simply and beauty that you can think, but if you need a solution in begginer level, you can try the following one:

As you have pointed out, we use the Squeeze theorem, we have that $$\lim_{n\to\infty}\sqrt[n]{2^n+3^n+4^n}\geq\lim_{n\to\infty}\sqrt[n]{4^n}=\lim_{n\to\infty}4=4.$$ By other hand, $$\lim_{n\to\infty}\sqrt[n]{2^n+3^n+4^n}=\lim_{n\to\infty}\sqrt[n]{4^n\left(\frac{2^n}{4^n}+\frac{3^n}{4^n}+1\right)}\leq\lim_{n\to\infty}\sqrt[n]{4^n\cdot3}=\lim_{n\to\infty}4\cdot3^{1/n}=4.$$ Hence $$4\leq\lim_{n\to\infty}\sqrt[n]{2^n+3^n+4^n}\leq4.$$ Finally we can conclude that $$\lim_{n\to\infty}\sqrt[n]{2^n+3^n+4^n}=4.$$