How do I factorize this expression? $$(2^n-3^n+n4^n)^{\frac{1}{n}}$$ so far I have: $$n4^n\left(\frac{1}{n} \left(\frac{1}{2}\right)^n-\frac{1}{n}\left(\frac{3}{4}\right)^n +1\right)^{\frac{1}{n}}$$
Forgot to mention the limits part of this question. How would I calculate the limit for this?
I don't see how this can be factored.
For the limit, if you take the $4^n$ out of the parens, the expression becomes $4(n-(3/4)^n+(1/2)^n)^{1/n}$ and this goes to 4 since $n^{1/n} \to 1$.