Can You Tackle Our Latest Math Challenge on Limits?

[anemone]

Here is this week's POTW:

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Evaluate the following limit:

\(\displaystyle \large \lim_{x \rightarrow 2} \left(\sqrt[6]{\frac{6x^4-12x^3-x+2}{x+2}} \times \frac{\sqrt[3]{x^3-\sqrt {x^2+60}}}{\sqrt{x^2-\sqrt[3] {x^2+60}}} \right)\)

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[anemone]

No one answered last week's problem. :(

You can find my solution below:

Spoiler

First, note that $6x^4-12x^3-x+2$ can be factorized as $6(2)^4-12(2)^3-(2)+2=96-96-2+2=0$, factorize it using the long polynomial division by the factor $x-2$, we get $6x^4-12x^3-x+2=(x-2)(6x^3-1)$.

So evaluating \(\displaystyle \large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^4-12x^3-x+2}{x+2}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)\) is equivalent in evaluating \(\displaystyle \large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{(x-2)(6x^3-1)}{x+2}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)\).

If we could think of another polynomial function of $f$, and multiply the top and bottom of the argument rational function inside the sixth root by the function of $f$, which for certain the function of $f$ also has a factor of $x-2$, then we could cross out the $x-2$ from top and bottom. But this function of $f$ must fulfill another criterion, it has to be closely related to $\sqrt[3]{x^3-\sqrt{x^2+60}}$ and $\sqrt{x^2-\sqrt[3]{x^2+60}}$:

\(\displaystyle \large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{(x-2)(6x^3-1)}{x+2}\cdot \frac{f(x)}{f(x)}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)\)

It's tempted to try out the option $f(x)=x^6-x^2-60$ since $f(2)=2^6-2^2-60=64-4-60=0$, and $f(x)=(x-2)(x+2)(x^4+4x^2+15)$

\(\displaystyle \large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^4-12x^3-x+2}{x+2}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)\)

\(\displaystyle =\large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{(x-2)(6x^3-1)}{x+2}\cdot \frac{f(x)}{f(x)}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)\)

\(\displaystyle =\large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{\cancel{(x-2)}(6x^3-1)}{x+2}\cdot \frac{x^6-x^2-60}{\cancel{(x-2)}(x+2)(x^4+4x^2+15)}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)\)

\(\displaystyle =\large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{x+2}\cdot \frac{x^6-x^2-60}{(x+2)(x^4+4x^2+15)}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)\)

\(\displaystyle =\large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{(x+2)^2(x^4+4x^2+15)}}\cdot \sqrt[6]{x^6-x^2-60}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)\)

\(\displaystyle =\large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{(x+2)^2(x^4+4x^2+15)}}\cdot \frac{\sqrt{x^6-x^2-60}}{\sqrt[3]{x^6-x^2-60}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)\,\,\small\text{as $\frac{1}{6}=\frac{1}{2}-\frac{1}{3}$}\)

\(\displaystyle =\large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{(x+2)^2(x^4+4x^2+15)}}\cdot \frac{\sqrt{(x^2)^3-\sqrt[3]{(x^2+60)^3}}}{\sqrt[3]{(x^3)^2-\sqrt{(x^2+60)^2}}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)\)

\(\displaystyle \small= \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{(x+2)^2(x^4+4x^2+15)}}\cdot \frac{\sqrt{(x^2-\sqrt[3]{x^2+60})}\sqrt{x^4+x^2\sqrt[3]{x^2+60}+\sqrt[3]{(x^2+60)^2}}}{\sqrt[3]{x^3-\sqrt{x^2+60}}\sqrt[3]{x^3+\sqrt{x^2+60}}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)\)

\(\displaystyle \small= \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{(x+2)^2(x^4+4x^2+15)}}\cdot \frac{\cancel{\sqrt{(x^2-\sqrt[3]{x^2+60})}}\sqrt{x^4+x^2\sqrt[3]{x^2+60}+\sqrt[3]{(x^2+60)^2}}}{\cancel{\sqrt[3]{x^3-\sqrt{x^2+60}}}\sqrt[3]{x^3+\sqrt{x^2+60}}}\cdot \frac{\cancel{\sqrt[3]{x^3-\sqrt{x^2+60}}}}{\cancel{\sqrt{x^2-\sqrt[3]{x^2+60}}}}\right)\)

\(\displaystyle = \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{(x+2)^2(x^4+4x^2+15)}}\cdot \frac{\sqrt{x^4+x^2\sqrt[3]{x^2+60}+\sqrt[3]{(x^2+60)^2}}}{\sqrt[3]{x^3+\sqrt{x^2+60}}}\right)\)

Now, we can safely evaluate the limit by plugging in $x=2$ into the expression and get:

\(\displaystyle \small \begin{align*}\lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^4-12x^3-x+2}{x+2}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)&=\lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{(x+2)^2(x^4+4x^2+15)}}\cdot \frac{\sqrt{x^4+x^2\sqrt[3]{x^2+60}+\sqrt[3]{(x^2+60)^2}}}{\sqrt[3]{x^3+\sqrt{x^2+60}}}\right)\\&=\sqrt[6]{\frac{6(2)^3-1}{(2+2)^2(2^4+4(2)^2+15)}}\cdot \frac{\sqrt{2^4+2^2\sqrt[3]{2^2+60}+\sqrt[3]{(2^2+60)^2}}}{\sqrt[3]{2^3+\sqrt{2^2+60}}}\\&=\sqrt[6]{\frac{47}{(4)^2(47)}}\cdot \frac{\sqrt{16+2^2(4)+(4)^2}}{\sqrt[3]{8+8}}\\&=\sqrt[6]{\frac{1}{4^2}}\cdot \frac{\sqrt{48}}{\sqrt[3]{16}}\\&=\frac{1}{2^{\frac{4}{6}}}\cdot \frac{4\sqrt{3}}{2^{\frac{4}{3}}}\\&=\frac{4\sqrt{3}}{4}\\&=\sqrt{3}\end{align*}\)