# I have one question about limits: it is required to find the limit <mstyle displaystyle="true" sc

I have one question about limits: it is required to find the limit $\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{{3}^{n}+{2}^{n}}$
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Isla Klein
Notice that
$\left({3}^{n}+{2}^{n}{\right)}^{1/n}={\left({5}^{n}\cdot \frac{{3}^{n}+{2}^{n}}{{5}^{n}}\right)}^{1/n}=5\cdot {\left[{\left(\frac{3}{5}\right)}^{n}+{\left(\frac{2}{5}\right)}^{n}\right]}^{1/n}.$
Thus, when you concluded that the second limit is 0 you virtually assumed that
$\underset{n\to \mathrm{\infty }}{lim}{\left[{\left(\frac{3}{5}\right)}^{n}+{\left(\frac{2}{5}\right)}^{n}\right]}^{1/n}=0,$
which is false. Your assumption is that since
$\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{3}{5}\right)}^{n}+{\left(\frac{2}{5}\right)}^{n}=0,$
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
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0.$
As such, your second calculation is simply incorrect. In general, if
$\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=0$
and
$\underset{n\to \mathrm{\infty }}{lim}{b}_{n}=0,$
then
$\underset{n\to \mathrm{\infty }}{lim}{{b}_{n}}^{{a}_{n}}$
can be equal to any nonnegative real number, or it can even not exist. You cannot simply conclude it to be 0.
###### Did you like this example?
babyagelesszj
The expression just equals:-
$\underset{n\to \mathrm{\infty }}{lim}3{\left(1+\left(\frac{2}{3}{\right)}^{n}\right)}^{\frac{1}{n}}$
Now this is ${1}^{\mathrm{\infty }}$ form.
You can use this whenever you run into this ${1}^{\mathrm{\infty }}$ form
$\underset{x\to a}{lim}f\left(x{\right)}^{g\left(x\right)}=\mathrm{exp}\left(\underset{x\to a}{lim}\left(f\left(x\right)-1\right)g\left(x\right)\right)$
when $\underset{x\to a}{lim}f\left(x{\right)}^{g\left(x\right)}$ is ${1}^{\mathrm{\infty }}$ form.
So using this we get it as :-
$3\mathrm{exp}\left(\underset{n\to \mathrm{\infty }}{lim}\frac{\left(\frac{2}{3}{\right)}^{n}}{n}\right)=3\mathrm{exp}\left(0\right)=3$
In general whenever you have something like
$\underset{n\to \mathrm{\infty }}{lim}{\left({a}_{1}^{n}+{a}_{2}^{n}+...+{a}_{m}^{n}\right)}^{\frac{1}{n}}$
where ${a}_{i}$ 's are distinct positive real numbers.You can use the above method and immediately write the answer as $\underset{1\le i\le m}{max}{a}_{i}$
That is we are strongly using the fact that ${a}^{n}\to 0$ when $0