limits (without using l'hopitals)

Question

kriyen · Answer

$$\lim_{x \rightarrow 8}      \frac{ x-8 }{ \sqrt[3]{x} - 2 }$$

kriyen · Answer

I get 16....but I'm not 100% sure...

anonymous · Answer

http://www.purplemath.com/modules/specfact2.htm Try using the difference of cubes formula. Let me know if that helps!

kirbykirby · Answer

See if you can view $x-8$ as a difference of cubes as they show on that page. Although they show $x^3-8$, you can use a similar procedure for $x-8$ [Although it's not perhaps a "true" difference of cubes, since you will get an expression with cube roots)

kirbykirby · Answer

Inspire yourself with the denominator :)

kriyen · Answer

Yeah I tried the difference of cubes way...I just want a confirmation of my answer...(16)

anonymous · Answer

is that a cube root? sorry, its just not clear... iam getting 12

kriyen · Answer

@Tushara How do you get 12?

anonymous · Answer

evaluate the limit for $$x \rightarrow 8^{+}$$ and $$x \rightarrow 8^{-}$$

anonymous · Answer

so substitute values, 8.001 and 7.999, both get u an answer close to 12 hence the limit exists and it is 12

kriyen · Answer

We need to evaluate without calculators...I'm not sure how to calculate the cube root of decimals without one.

anonymous · Answer

i see

kriyen · Answer

Okay I figured it out...I made a mistake when solving the difference of cube

$$\lim_{x \rightarrow 8} \frac{ x - 8 }{ \sqrt[3]{x} -2 }$$
$$= \frac{ (\sqrt[3]{x} - 2)((\sqrt[3]{x})^2 + (2)(\sqrt[3]{x}) + (2)^2)}{ \sqrt[3]{x}-2 }$$
$$= 4 + 4 + 4 $$
$$= 12$$

kriyen · Answer

@Tushara @cormacpayne @kirbykirby  Thanks for the help =)

kirbykirby · Answer

:)