Question

Can someone explain how to get the answer in the picture from the problem in the picture? It's in the L'Hopital's rule section of our calculus homework

http://www3.wolframalpha.com/Calculate/MSP/MSP98421a3gd7f680a0833a000069a0d634d37c1ff4?MSPStoreType=image/gif&s=46&w=132&h=55

Answer 1

I'm tempted to rewrite the large numerator with a common denominator for the two tiny fractions. After doing the same to the large denominator, I'm guessing something will pop out at us.

Answer 2

It's equivalent to this: (x^(4/3)+12x^2)/(8x^(5/3)+2x^2)
http://www3.wolframalpha.com/Calculate/MSP/MSP4041a3gdf06430bi9b800003fec573ge7ie4aef?MSPStoreType=image/gif&s=37&w=83&h=40

which Wolfram Alpha gives an explanation for, although that explanation is not very helpful

Answer 3

2 > 4/3 and 2 > 5/3 This is almost done. What would you do with this? \[\frac{x^{2} + 12x^{3}}{8x + 2x^{3}}\]

Answer 4

now divide everything by x^2 ...

Answer 5

Or just multiply top and bottom of the equation by x^(4/3). You have terms that go to zero and a quotient of 12/2 left.

Answer 6

Oh, I think I got it--one second, let me properly work it out

Answer 7

\[\large\lim_{x\rightarrow\infty}\frac{\frac{1}{x^{2}}+\frac{12}{x^{\frac{4}{3}}}}{\frac{8}{x^{\frac{5}{3}}}+\frac{2}{x^{\frac{4}{3}}}}*\frac{x^{\frac{4}{3}}}{x^{\frac{4}{3}}}\]\[\large\lim_{x\rightarrow\infty}\frac{\frac{1}{x^{\frac{2}{3}}}+12}{\frac{8}{x^{\frac{1}{3}}}+2}=\frac{0+12}{0+2}\]

Answer 8

Well, except for the fact that you substituted and this totally mistreats the concept of the limit, I think you have it.