Limit of exponential series Find the following limit:
$$\lim_{n \rightarrow \infty} \sqrt[n]{ \left( 3^n-2^n+1/2^n \right ) x^n }$$
How to calculate this limit? I am lookin for the solution that would involve use of squeeze theorem.
First note that 
$$ 3x \sqrt[n]{1 - \left(\frac 23\right)^n + \left(\frac 16\right)^n} = \sqrt[n]{(3^n - 2^n + 1/2^n)x^n}. $$
Now 
$$ 1 - \left(\frac 23\right)^n + \left(\frac 16\right)^n \le 1 $$
and, as the left hand side tends to one, finally (that is for all large enough $n$):
$$ \frac 1n \le 1 - \left(\frac 23\right)^n + \left(\frac 16\right)^n $$
That gives, by the squeeze theorem, that 
$$ 1 = \lim_n \frac 1{\sqrt[n]n} \le \lim_n \sqrt[n]{1 - \left(\frac 23\right)^n + \left(\frac 16\right)^n} \le \lim_n 1 = 1 $$
Hence
$$ \lim_n\sqrt[n]{(3^n - 2^n + 1/2^n)x^n} = 3x $$
Hint
$$\lim_{n\to\infty}(a^n_{1}+a^n_{2}+\cdots+a^n_{m})^{1/n}=\max{(a_{1},a_{2},\cdots,a_{m})}$$