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?

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 $0.$
