Question about the limit $\displaystyle\lim_{n\rightarrow \infty}\sqrt[n]{3^n+ 2^n}$ I have one question about limits: it is required to find the limit $\displaystyle\lim_{n \rightarrow \infty}\sqrt[n]{3^n + 2^n}$.
I calculated it like this:
$$(3^n+2^n)^{1/n} = \left(3^n \cdot \frac{3^n+2^n}{3^n}\right)^{1/n} \rightarrow (3^n)^{1/n} = 3.$$
but what if I divide, say by $5^n$:
$$\left(5^n \cdot \frac{3^n+2^n}{5^n}\right)^{1/n} \rightarrow (5^n \cdot 0)^{1/n} = 0.$$
Why is the second solution wrong?
 A: Notice that $$(3^n+2^n)^{1/n}=\left(5^n\cdot\frac{3^n+2^n}{5^n}\right)^{1/n}=5\cdot\left[\left(\frac35\right)^n+\left(\frac25\right)^n\right]^{1/n}.$$ Thus, when you concluded that the second limit is $0,$ you virtually assumed that $$\lim_{n\to\infty}\left[\left(\frac35\right)^n+\left(\frac25\right)^n\right]^{1/n}=0,$$ which is false. Your assumption is that since $$\lim_{n\to\infty}\left(\frac35\right)^n+\left(\frac25\right)^n=0,$$ that the former limit must also be $0.$ However, this is not the case: in the cases where the exponent has limit $0,$ it is not sufficient for the base to have limit $0$ in order for the entire power to have limit $0.$ The exponent is $1/n,$ and notice that $$\lim_{n\to\infty}\frac1{n}=0.$$ As such, your second calculation is simply incorrect. In general, if $$\lim_{n\to\infty}a_n=0$$ and $$\lim_{n\to\infty}b_n=0,$$ then $$\lim_{n\to\infty}{b_n}^{a_n}$$ can be equal to any nonnegative real number, or it can even not exist. You cannot simply conclude it to be $0.$
A: A separate approach:-
The expression just equals:-
$$\lim_{n\to\infty}3\left(1+(\frac{2}{3})^{n}\right)^{\frac{1}{n}}$$.
Now this is $1^{\infty}$ form.
You can use this whenever you run into this $1^{\infty}$ form
$$\lim_{x\to a}f(x)^{g(x)} = \exp(\lim_{x\to a} (f(x)-1)g(x))$$ when $\lim_{x\to a}f(x)^{g(x)}$ is $1^{\infty}$ form.
So using this we get it as :-
$$3\exp(\lim_{n\to\infty}\frac{(\frac{2}{3})^{n}}{n})=3\exp(0)=3$$.
In general whenever you have something like
$$\lim_{n\to\infty}\left(a_{1}^{n}+a_{2}^{n}+...+a_{m}^{n}\right)^{\frac{1}{n}}$$. where $a_{i}$'s are distinct positive real numbers.You can use the above method and immediately write the answer as :-  $\displaystyle\max_{1\leq i\leq m}a_{i}$.
That is we are strongly using the fact that $a^{n}\to 0$ when $0<a<1$.
