Evaluating limits using change of variable/ variable substitution

**Bri22** · February 27th 2013, 06:38 PM

Hi there! I am currently in grade 12, studying calculus 30 which is just calculus. I have a question about the above topic posted. One contains roots and I'm not sure exactly how to solve it. Any help would be great!

1) lim ( X^1/3 - 3) / (x - 27)
X-->27

2) lim. (square root of x then minus 2 outside of square root / square root x^3 - 8 ( 8 is not under the root)
x--> 4

**TheEmptySet** · February 27th 2013, 06:56 PM

Use the difference of cubes formula to reduce each fraction $a^3-b^3=(a-b)(a^2+ab+b^2)$

For the first one we get

$x-27=(x^{\frac{1}{3}})^3-3^3=(x^{\frac{1}{3}}-3)(x^{\frac{2}{3}}+3x^{\frac{1}{3}}+9)$

This will reduce out with the factor of the numerator and now the function is continuous so you can just evaluate at x=27

$\lim_{x \to 27}\frac{x^{\frac{1}{3}}-3)}{(x^{\frac{1}{3}}-3)(x^{\frac{2}{3}}+3x^{\frac{1}{3}}+9)}=\lim_{x \to 27}\frac{1}{(x^{\frac{2}{3}}+3x^{\frac{1}{3}}+9)}= \frac{1}{27}$

**Prove It** · February 27th 2013, 10:22 PM

Originally Posted by Bri22

Hi there! I am currently in grade 12, studying calculus 30 which is just calculus. I have a question about the above topic posted. One contains roots and I'm not sure exactly how to solve it. Any help would be great!

1) lim ( X^1/3 - 3) / (x - 27)
X-->27

2) lim. (square root of x then minus 2 outside of square root / square root x^3 - 8 ( 8 is not under the root)
x--> 4

$\displaystyle \begin{align*} \lim_{x \to 4} \frac{\sqrt{x} - 2}{\sqrt{x^3} - 8} &= \lim_{x \to 4} \frac{\sqrt{x} - 2}{\left( \sqrt{x} \right)^3 - 2^3} \\ &= \lim_{x \to 4} \frac{\sqrt{x} - 2}{\left( \sqrt{x} - 2 \right) \left( x + 2\sqrt{x} + 4 \right) } \\ &= \lim_{x \to 4} \frac{1}{x + 2\sqrt{x} + 4} \\ &= \frac{1}{4 + 2\sqrt{4} + 4} \\ &= \frac{1}{12} \end{align*}$

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