Find the limit of $a_n=\frac{\sqrt[n]{n^n +cn}}{cn+c/(n+1)}$ with $c>0$.

Question

Find the limit of $a_n=\frac{\sqrt[n]{n^n +cn}}{cn+c/(n+1)}$ with $c>0$.

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I'm sorry to post such a (probably) trivial problem, but I'm just not seeing it. My best shot at it was:

$$\frac{\sqrt[n]{n^n +cn}}{cn+c/(n+1)}=\frac{\sqrt[n]{n^{n-1} +c}\cdot\sqrt[n]{n}}{c(n+\frac{1}{n+1})}\longrightarrow\frac{1\cdot\infty}{\infty}=\infty.$$

I have officially no knowledge of calculus yet, meaning no Taylor series expansion.

Answer 1

$$ a_n=\frac{\sqrt[n]{n^n +cn}}{cn+c/(n+1)} = \frac{(n+1)n\sqrt[n]{1+c n^{-(n-1)}}}{c((n+1)n+1)} = \frac{\sqrt[n]{1+c n^{-(n-1)}}}{c(1+\frac{1}{(n+1)n})} $$

hence

$$ \lim_{n\to \infty}a_n = \frac{1}{c} $$