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Evaluate the limit

$$ \lim_{x\rightarrow \infty}\left(2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right) $$

My Attempt:

To simplify notation, let $A = \left(\sqrt[3]{x^3+x^2+1}\right)$ and $B = \left(\sqrt[3]{x^3-x^2+1}\right)$. Now

$$ \begin{align} 2x^2 &= A^3-B^3\\ x &= \sqrt{\frac{A^3-B^3}{2}} \end{align} $$

So the limit becomes

$$\lim_{x\rightarrow \infty}\left(\sqrt{\frac{A^3-B^3}{2}}-A-B\right)$$

How can I complete the solution from this point?

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  • $\begingroup$ Factor x out of the roots and write the expression as x*(2-something-something). That may help. $\endgroup$ – Johannes Ernst Jun 30 '14 at 3:34
  • $\begingroup$ $\sqrt[3]{x^3+x^2+1}-\sqrt[3]{x^3-x^2+1}=2x-\cfrac{2x^{-1}}{9}+o(x^{-1})$ $\endgroup$ – Fabien Jun 30 '14 at 3:48
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Note that $\sqrt[3]{x^3+x^2+1}=x\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^3}}$, with a similar expression for the other one. Make the substitution $t=1/x$. Our expression becomes $$\frac{2-\sqrt[3]{1+t+t^3}-\sqrt[3]{1-t+t^3}}{t},$$ and we want to find the limit of this as $t$ approaches $0$ from the right. Note that the conditions for using L'Hospital's Rule are met. So our limit is $$\lim_{t\to 0^+} -\frac{1}{3}(1+3t^2)(1+t+t^3)^{-2/3}-\frac{1}{3}(-1+3t^2)(1-t+t^3)^{-2/3}.$$ This one is easy.

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It is easier to break into two limits, $x-\sqrt[3]{x^3+x^2+1}$ and $x-\sqrt[3]{x^3-x^2+1}$ then we shall see the first limits to $-\frac{1}{3}$ and the second $\frac{1}{3}$ so the sum limits to zero.

Note the more general result,

$$x-\sqrt[n]{x^n +ax^{n-1}+\cdots}\rightarrow -\frac{a}{n}$$ If we write $A=\sqrt[n]{x^n +ax^{n-1}+\cdots}$ then

$$x-A=\frac{x^n-A^n}{x^{n-1}+x^{n-2}A +\cdots +A^{n-1}}=\frac{-ax^{n-1}+\cdots}{x^{n-1}+x^{n-2}A +\cdots +A^{n-1}}$$

and since $\frac{A}{x}\rightarrow 1$ the rest is easy.

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Hint

Since $x$ goes to infinity, let us write $$\sqrt[3]{x^3+x^2+1}=x\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}}$$ and let us define $y=\frac{1}{x}+\frac{1}{x^2}$. Now, look at the Taylor expansion of $$\sqrt[3]{1+y}=1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)$$ and replace $y$ by its definition and expand to get finally $$\sqrt[3]{x^3+x^2+1}=x \Big(1+\frac{1}{3 x}-\frac{1}{9 x^2}+...\Big)\simeq x-\frac{1}{9 x}+\frac{1}{3}$$ Similarly, the same method should give $$\sqrt[3]{x^3-x^2+1} \simeq x-\frac{1}{9 x}-\frac{1}{3}$$

I am sure that you can take from here.

More intuitively, you could have noticed that, since $x$ goes to $\infty$,$$\sqrt[3]{x^3 \pm x^2+1}=x\sqrt[3]{1 \pm\frac{1}{x}+\frac{1}{x^2}}$$ behaves just as $x$ and then the limit of the expression is $0$. However, this approach allows you to show "how" the expression goes to $0$ (by positive or negative values).

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$a+b=\dfrac{a^3+b^3}{a^2-ab+b^2}$
I think this identity can be used to simplify your expression.
Let $a=\sqrt[3]{x^3+x^2+1}$ and $b=\sqrt[3]{x^3-x^2+1}.$
Then
$a+b=\dfrac{(x^3+x^2+1)+(x^3-x^2+1)}{(x^3+x^2+1)^{2/3}-(x^3+x^2+1)^{1/3}(x^3+x^2+1)^{1/3}+(x^3+x^2+1)^{2/3}}\\=\dfrac{2(x+1/x^2)}{(1+1/x+1/x^3)^{2/3}-(1+1/x+1/x^3)^{1/3}(1+1/x+1/x^3)^{1/3}+(1+1/x+1/x^3)^{2/3}}$
As $x\rightarrow \infty$ we have $a+b\rightarrow 2x.$ Hence $$\lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}\\= \lim_{x\rightarrow \infty} {2x-(a+b)}=0$$

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    $\begingroup$ This still needs some minor corrections, although basic idea seems to be ok. You cannot say that $a(x)+b(x)\to 2x$ when $x\to\infty$. (Since a and b depend on x.) $\endgroup$ – Martin Sleziak Aug 21 '14 at 8:49
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As $x \to\infty$, $A\to x$ and $B\to x$. Therefore the expression would tend to $0$.

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    $\begingroup$ This kind of reasoning is faulty for example $x-\sqrt[3]{x^3+x^2+1} \rightarrow -\frac{1}{3}$ $\endgroup$ – Rene Schipperus Jun 30 '14 at 4:08

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