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Evaluate $\displaystyle{\lim_{x \to \infty} (x^3+6x^2+1)^{\frac13}-(x^2}+x+1)^{\frac12}$

I tried writing it as $$\lim_{x \to \infty}\frac{x^3+6x^2+1}{(x^3+6x^2+1)^{\frac23}}-\frac{x^2+x+1}{(x^2+x+1)^{\frac12}}$$ but I do not know how to continue from this point.

I really appreciate your time and effort. Thank you very much!

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    $\begingroup$ Hint: rewrite as$$\lim_{x\to\infty}((x^3+6x^2+1)^{1/3}-x)-\lim_{x\to\infty}((x^2+x+1)^{1/2}-x)$$then use$$a-b=\frac{a^n-b^n}{\sum_{k=0}^{n-1}a^kb^{n-1-k}}$$to evaluate each limit. $\endgroup$
    – J.G.
    Nov 11, 2023 at 9:47
  • $\begingroup$ Thank you very much! $\endgroup$
    – george_o
    Nov 11, 2023 at 9:49

2 Answers 2

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Let $t=\frac{1}{x}$. Then

$$(x^3+6x^2+1)^{\frac13}-(x^2+x+1)^{\frac12} = \frac{(t^3+6t+1)^{\frac13}-(t^2+t+1)^{\frac12}}{t} = \frac{f(t)-f(0)}{t-0} $$

where $f(t)=(t^3+6t+1)^{\frac13}-(t^2+t+1)^{\frac12}$. Therefore, if it exists, the limit you are looking for is equal to $\lim_{t\to 0^+}\frac{f(t)-f(0)}{t-0}$. Since $f$ is differentiable at $0$, the limit exists and is equal to $f'(0)$.

Using Taylor series, $f(t)=0+\frac{3}{2}t+...$, so $f'(0)=\frac{3}{2}$.

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  • $\begingroup$ Thank you very much! You really helped me $\endgroup$
    – george_o
    Nov 11, 2023 at 9:49
  • $\begingroup$ Looks way overkill. Also how do you compute the Taylor series of $f$? $\endgroup$
    – Adayah
    Nov 11, 2023 at 19:26
  • $\begingroup$ @Adayah: you only need a few terms of the Taylor series, so it is quite easy (it is similar to what Sine of the Time did). You can also compute the derivative of $f$ using usual formulas if you prefer. $\endgroup$
    – Taladris
    Nov 12, 2023 at 3:15
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$$\lim_{x \to \infty} (x^3+6x^2+1)^{\frac13}-(x^2+x+1)^{\frac12}$$ Using $(1+t)^a\sim1+at $ as $t\to0$, you have: $$\sqrt[3]{x^3+6x^2+1}-\sqrt{x^2+x+1}=x\sqrt[3]{1+6/x+1/x^3}-x\sqrt{1+1/x+1/x^2}\sim \\ x(1+2/x)-x(1+1/2x)=2-1/2=3/2$$

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  • $\begingroup$ What exactly do you mean by $\sim$ ? $\endgroup$
    – Adayah
    Nov 11, 2023 at 19:28
  • $\begingroup$ @Adayah asymptotic equivalence $\endgroup$ Nov 11, 2023 at 19:28
  • $\begingroup$ If you mean $f \sim g \iff \lim_{x \to \infty} \frac{f(x)}{g(x)} = 1$, then the argument is incorrect for multiple reasons. A typical example of how this fails: $(x+1) - x \sim x - x = 0$, so $\lim_{x \to \infty} (x+1) - x = 0$. $\endgroup$
    – Adayah
    Nov 11, 2023 at 19:37
  • $\begingroup$ @Adayah (x+1)-x =1 so it doesn't make sense saying $\sim 0$ $\endgroup$ Nov 11, 2023 at 19:51
  • $\begingroup$ It doesn't, right? But I only followed the same rules that you did. How do you rigorously justify this? $$x \sqrt[3]{1+6/x+1/x^3} - x \sqrt{1+1/x+1/x^2} \sim x(1+2/x) - x(1+1/2x)$$ $\endgroup$
    – Adayah
    Nov 11, 2023 at 20:32

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