Finding limit function $\lim_{n \rightarrow \infty} n ((x^2 +x + 1)^{1/n} -1)$ \begin{align}
f(x) &= \lim_{n \rightarrow \infty} n ((x^2 +x + 1)^{1/n} -1) \\&= \lim_{n \rightarrow \infty} n ((\infty)^{1/n} -1) \\&= \lim_{n \rightarrow \infty} n (1 -1)\\& = 
 \lim_{n \rightarrow \infty} n \cdot 0 \\&= 0 
\end{align}
Did I solve it correctly??
 A: Put $\,a:=x^2+x+1\,$ . Note that $\,x\in\Bbb R\implies a>0\;$ (why?) . Thus, you want 
$$\lim_{n\to\infty} n\left(\sqrt[n] a-1\right)$$
Let us define for a continuous variable
$$x>0\;,\;\;f(x):= x(\sqrt[x]a-1)=\frac{\sqrt[x]a-1}{\frac1x}$$
Now, you can apply l'Hospital when $\,x\to\infty\,$ (why?) , so
$$\lim_{x\to\infty}f(x)\stackrel{\text{l'H}}=\frac{-\frac1{x^2}a^{\frac1x}\log a}{-\frac1{x^2}}=\lim_{x\to\infty}\frac{a^{1/x}\log a}1=\log a$$
Thus....
A: Note that $$\lim\limits_{n \rightarrow \infty} n ((x^2 +x + 1)^{\tfrac{1}{n}} -1) =\lim\limits_{n\rightarrow \infty}  \dfrac{(x^2 +x + 1)^{\tfrac{1}{n}} -1}{\dfrac{1}{n}}. $$
Denoting $h(x)=x^2 +x + 1,$ we have
$$\lim\limits_{n\rightarrow \infty}  \dfrac{(x^2 +x + 1)^{\tfrac{1}{n}} -1}{\dfrac{1}{n}}=\lim\limits_{n\rightarrow \infty}  \dfrac{(h(x))^{\tfrac{1}{n}} -1}{\dfrac{1}{n}}=\ln{h(x)}.$$
A: Hint: Temporarily, call $x^2+x+1$ by the name $b$. Then 
$$(x^2+x+1)^{1/n}=b^{1/n}=e^{(\ln b)/n}.$$ Let $h=\frac{1}{n}$. Then our expression can be rewritten as
$$\frac{e^{(\ln b)h}-1}{h}.$$
Let $h$ approach $0$. Note that
$$\lim_{h\to 0}  \frac{e^{(\ln b)h}-1}{h}$$
is the derivative of the function $g(t)=e^{(\ln b)t}$ at $t=0$.
The derivative of $e^{(\ln b)t}$ is $(\ln b)e^{(\ln b)t}$. Now we are close to the end.
