Finding limit function $\lim_{n \rightarrow \infty} n ((x^2 +x + 1)^{1/n} -1)$

Question

\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??

Answer 1

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....

Answer 2

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 $$\lim\limits_{n\rightarrow \infty} \dfrac{(x^2 +x + 1)^{\tfrac{1}{n}} -1}{\dfrac{1}{n}}=\lim\limits_{n\rightarrow \infty} \dfrac{(h(x))^{\tfrac{1}{n}} -1}{\dfrac{1}{n}}=\ln{h(x)}.$$

Answer 3

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.