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\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??

Martin Sleziak
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Stas
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    How, why did you pass from $,x^2+x+1,$ to $;\infty;$ ?? And how did you deduce that $,\infty^{1/n}=1;$ , whatever may "infinite raised to some power" mean...? And how would you justify passing to the limit when $,n\to\infty,$ from the second to the third expression above, yet the exterior $\;n;$ still remains?! – DonAntonio Jun 30 '13 at 18:03
  • See this post. – David Mitra Jun 30 '13 at 18:04
  • @Mitra, thanks for the link – Stas Jun 30 '13 at 18:12

3 Answers3

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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....

DonAntonio
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  • $a>0$ because this polynomial doesn't intersect x-axis. That's why I reduce it to infinity. My reasoning was based on $\lim \sqrt[n]{a}=1$. And, as I thought, no matter what value the polynomial equals because in the end $\lim \sqrt[n]{a}=1$. – Stas Jun 30 '13 at 18:15
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    Yes, but then $,a=x^2+x+1>0,$ because the expression in $,x,$ is an upwards parabola, but this is not $,\infty,$ ! – DonAntonio Jun 30 '13 at 18:17
  • I found another my mistake: $\lim n (\sqrt[n]{a}-1) \neq \lim n \cdot \lim (\sqrt[n]{a}-1) $. Right? But I thought that the nice thing about limits is that one can break expression into smaller parts, like $\lim f\pm g = \lim f \pm \lim g$; $\lim f \cdot g = \lim f \cdot \lim g$ – Stas Jun 30 '13 at 18:27
  • You are right! $;\lim a_nb_n=\lim a_n\cdot\lim b_n;$ only when both limits exist separatedly and they are finite – DonAntonio Jun 30 '13 at 18:46
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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 $$\lim\limits_{n\rightarrow \infty} \dfrac{(x^2 +x + 1)^{\tfrac{1}{n}} -1}{\dfrac{1}{n}}=\lim\limits_{n\rightarrow \infty} \dfrac{(h(x))^{\tfrac{1}{n}} -1}{\dfrac{1}{n}}=\ln{h(x)}.$$

M. Strochyk
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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.

André Nicolas
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