MHB Limit of (2 - sqrt(2 + x)) / (2^(1/3) - (4 - x)^(1/3)) at x = 2

[Jameson]

Evaluate the following limit without using calculus:

$$\lim_{x\rightarrow 2}\frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}}$$
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[Jameson]

Congratulations to the following members for their correct solutions:

1) MarkFL
2) anemone
3) BAdhi
4) Sudharaka

Honorable mention goes to kaliprasad.

Solution (from anemone):

Spoiler

We're asked to find the limit of $\displaystyle \lim_{x\to 2} \frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^{\frac{1}{3}}}$.

Notice that the expressions of both denominator and numerator approach zero as x approaches 2, this means that the given limit cannot be derived by intuition but we notice too that the function can be simplified further.

Using the formula $\displaystyle (a^3-b^3)=(a-b)(a^2+ab+b^2)$ to rationalize the denominator, we get:

$\displaystyle \lim_{x\to 2} \frac{(2-\sqrt{2+x}) (2^{\frac{2}{3}}+(2^{\frac{1}{3}})((4-x)^{\frac{1}{3}})+(4-x)^{\frac{2}{3}})}{(2^{\frac{1}{3}}-(4-x)^{\frac{1}{3}}) (2^{\frac{2}{3}}+(2^{\frac{1}{3}})((4-x)^{\frac{1}{3}})+(4-x)^{\frac{2}{3}})}$

$\displaystyle= \lim_{x\to 2} \frac{(2-\sqrt{2+x}) (2^{\frac{2}{3}}+(8-2x)^{\frac{1}{3}}+(4-x)^{\frac{2}{3}})}{(2^{\frac{1}{3}})^3-((4-x)^{\frac{1}{3}})^3}$

$\displaystyle= \lim_{x\to 2} \frac{(2-\sqrt{2+x}) (2^{\frac{2}{3}}+(8-2x)^{\frac{1}{3}}+(4-x)^{\frac{2}{3}})}{2-(4-x)}$

$\displaystyle= \lim_{x\to 2} \frac{(2-\sqrt{2+x}) (2^{\frac{2}{3}}+(8-2x)^{\frac{1}{3}}+(4-x)^{\frac{2}{3}})}{x-2}$

Now we know in our next step we need to get rid of the first factor in the numerator because $\displaystyle 2-\sqrt{2+x}$ approaches zero as x approaches 2. We can do that by multiplying the top and the bottom of the function by $\displaystyle 2+\sqrt{2+x}$, i.e.

$\displaystyle \lim_{x\to 2} \frac{(2-\sqrt{2+x}) (2^{\frac{2}{3}}+(8-2x)^{\frac{1}{3}}+(4-x)^{\frac{2}{3}})}{x-2}$

$\displaystyle = \lim_{x\to 2} \frac{(2-\sqrt{2+x}) (2^{\frac{2}{3}}+(8-2x)^{\frac{1}{3}}+(4-x)^{\frac{2}{3}})( 2+\sqrt{2+x})}{(x-2)( 2+\sqrt{2+x})}$

$\displaystyle = \lim_{x\to 2} \frac{(2-x) (2^{\frac{2}{3}}+(8-2x)^{\frac{1}{3}}+(4-x)^{\frac{2}{3}})}{(x-2)( 2+\sqrt{2+x})}$

$\displaystyle = \lim_{x\to 2} \frac{(2-x) (2^{\frac{2}{3}}+(8-2x)^{\frac{1}{3}}+(4-x)^{\frac{2}{3}})}{-(2-x)( 2+\sqrt{2+x})}$

$\displaystyle = \lim_{x\to 2} \frac{\cancel{(2-x)} (2^{\frac{2}{3}}+(8-2x)^{\frac{1}{3}}+(4-x)^{\frac{2}{3}})}{-\cancel{(2-x)}( 2+\sqrt{2+x})}$

Since we are now left with a closed form function that is defined when x = 2, we can now evaluate the limit by substitution:

$\displaystyle \lim_{x\to 2} \frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^{\frac{1}{3}}}$

$\displaystyle = \lim_{x\to 2} \frac{(2^{\frac{2}{3}}+(8-2x)^{\frac{1}{3}}+(4-x)^{\frac{2}{3}})}{-( 2+\sqrt{2+x})}$

$\displaystyle = \frac{2^{\frac{2}{3}}+4^{\frac{1}{3}}+2^{\frac{2}{3}}}{-( 2+\sqrt{2+2})}$

$\displaystyle = \frac{3(2)^{\frac{2}{3}}}{-4}$

$\displaystyle = -\frac{3}{2(2)^{\frac{1}{3}}}$