Limit with unknown parameter What possible values can take $a,b\in\mathbb{R}$ such that
$$
\lim\limits_{n\rightarrow\infty}\left(  \sqrt[3]{an^{3}+bn^{2}+1}-\log
_{5}\left(  3^{n}+4^{n}+5^{n}\right)  -\sqrt[4]{n^{4}+1}\right)  =1~?
$$
Denote $\left(  a_{n}\right)  $ the expresion inside limit. My idea is to
manipulate the expresions one-by-one inside the limit and I get that
\begin{align*}
\lim\limits_{n\rightarrow\infty}\left(  a_{n}-1\right)    & =\lim
\limits_{n\rightarrow\infty}\left(  \sqrt[3]{an^{3}+bn^{2}+1}-1-\log_{5}%
5^{n}\left(  \left(  \dfrac{3}{5}\right)  ^{n}+\left(  \dfrac{4}{5}\right)
^{n}+1\right)  -\sqrt[4]{n^{4}+1}\right)  \\
& =\lim\limits_{n\rightarrow\infty}\left(  \sqrt[3]{an^{3}+bn^{2}+1}%
-1-n+\log_{5}\left(  \left(  \dfrac{3}{5}\right)  ^{n}+\left(  \dfrac{4}%
{5}\right)  ^{n}+1\right)  -\sqrt[4]{n^{4}+1}\right)  \\
& =\lim\limits_{n\rightarrow\infty}\left(  \sqrt[3]{an^{3}+bn^{2}%
+1}-1-n-\sqrt[4]{n^{4}+1}\right)  \\
& =\lim\limits_{n\rightarrow\infty}\left(  \sqrt[3]{an^{3}+bn^{2}%
+1}-1-2n-\left(  \sqrt[4]{n^{4}+1}-n\right)  \right)  \\
& =\lim\limits_{n\rightarrow\infty}\left(  \dfrac{an^{3}+bn^{2}+1-1}
{\sqrt[3]{an^{3}+bn^{2}+1}^{2}+\sqrt[3]{an^{3}+bn^{2}+1}+1}-2n-\lim
\limits_{n\rightarrow\infty}\dfrac{1}{n^{\alpha}+...}\right)  \\
& =\lim\limits_{n\rightarrow\infty}\left(  \dfrac{an^{3}+bn^{2}}
{\sqrt[3]{an^{3}+bn^{2}+1}^{2}+\sqrt[3]{an^{3}+bn^{2}+1}+1}-2n\right)  .
\end{align*}
where I have used $\log_{5}\left(  \left(  \dfrac{3}{5}\right)  ^{n}+\left(
\dfrac{4}{5}\right)  ^{n}+1\right)  \rightarrow\log_{5}1=0.$
The limit $\lim\limits_{n\rightarrow\infty}\left(  a_{n}-1\right)  $ should be
equal with $0$, but I do not think that my last relation would imply this
thing. Am I wrong something?
 A: Write:
$$\left\{\begin{array}{lcl}
\sqrt[3]{an^{3}+bn^{2}+1}&=& n\, \sqrt[3]{a+\frac{b}n+\frac1{n^2}}\\
\log_{5}\left(  3^{n}+4^{n}+5^{n}\right) &=& 
%\log_{5}\left( 5^n\left({\left(\frac35\right)}^{n}+{\left(\frac45\right)}^{n}+1\right)\right)=
n + \log_{5}\left( {\left(\frac35\right)}^{n}+{\left(\frac45\right)}^{n}+1\right)\\
\sqrt[4]{n^{4}+1} &=& n\,\sqrt[4]{1+\frac1{n^4}}
\end{array}\right.
.$$
Then, we have
\begin{align}
a_n
&=
n\, \sqrt[3]{a+\frac{b}n+\frac1{n^2}}
-n -\log_{5}\left( {\left(\frac35\right)}^{n}+{\left(\frac45\right)}^{n}+1\right)
-n\,\sqrt[4]{1+\frac1{n^4}}
\\&=
n\left(\sqrt[3]{a+\frac{b}n+\frac1{n^2}} - 1 - \sqrt[4]{1+\frac1{n^4}}\right)
-\frac1n\,\log_{5}\left( {\left(\frac35\right)}^{n}+{\left(\frac45\right)}^{n}+1\right).
\end{align}
Now, as $n\to\infty$, we have that
$$\frac1n\,\log_{5}\left( {\left(\frac35\right)}^{n}+{\left(\frac45\right)}^{n}+1\right) \to 0.$$
Moreover, as $n\to\infty$ we have 
$$\sqrt[3]{a+\frac{b}n+\frac1{n^2}} - 1 - \sqrt[4]{1+\frac1{n^4}}
\to \sqrt[3]a - 2.$$
Hence, if $a > 8$, then $a_n\to+\infty$ and if $a < 8$, then $a_n\to-\infty$.
If $a=8$, then we have an indeterminate.
Consider
$$f(x) = 
\frac{\displaystyle
\sqrt[3]{8+\frac{b}x+\frac1{x^2}} - 1 - \sqrt[4]{1+\frac1{x^4}}
}{1/x}.$$
As $x\to\infty$, $f(x)$ is an indeterminate expression of the type $0/0$.
We apply L'Hôpital, obtaining that
\begin{align}
\lim_{x\to\infty}f(x)
&=
\lim_{x\to\infty}
\frac{\displaystyle
-\frac{b}{3x^2{\left(8+\frac{b}x+\frac1{x^2}\right)}^{2/3}}
-\frac{2}{3x^3{\left(8+\frac{b}x+\frac1{x^2}\right)}^{2/3}}
+\frac{1}{x^5{\left(1+\frac1{x^4}\right)}^{3/4}}
}{-1/x^2}
\\&=\lim_{x\to\infty}
\frac{b}{3{\left(8+\frac{b}x+\frac1{x^2}\right)}^{2/3}}
+\frac{2}{3x{\left(8+\frac{b}x+\frac1{x^2}\right)}^{2/3}}
-\frac{1}{x^3{\left(1+\frac1{x^4}\right)}^{3/4}}
\\&=
\frac{b}{3\cdot 8^{2/3}} = \frac b{12}.
\end{align}
It follows that
$$\lim_{n\to\infty}a_n
=\left\{\begin{array}{lccr}
-\infty&&&a<8\\
b/12&&&a=8\\
+\infty&&&a>8
\end{array}
\right.,$$
so $\lim_{n\to\infty}a_n = 1$ if and only if $(a,b) = (8, 12)$.
