What is the limit of $\lim_{x \to \infty}\sqrt[x]{2^x+3^x+4^x}$ 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?
 A: 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:
$$\lim_{n\to\infty}\left(\sqrt[n]{2^{n}+3^{n}+4^{n}}\right)=\|\vec{x}\|_{\infty}=\max\{2,3,4\}=4$$
A: Hint: Factor $4^x$ inside the radical.
A: The dominating term in the radical is $4^x$. So, we have $L=4^{(x/x)} = 4$
A: 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
$$\lim_{n\to\infty}\sqrt[n]{2^n+3^n+4^n}=4.$$
