find the limit x→ ∞, [x + x ^(1/2) + x ^(1/3)] / [x^(2/3) + x^(1/4) ]

[hndalama]

[x + x^1/2 + x^1/3] / [x^2/3 + x^1/4 ]

I determine that the numerator and the denominator both approach ∞ so LHopital's rule can be used. when I use lhopital rule I get
[1 + 0.5x^-0.5 +1/(3x^2/3) ] / [2/(3x^1/3) + 1/(4x^3/4) ] . In the numerator, i determine that the term with 1 over a value with x will go to 0 so the numerator becomes just 1, in the denominator i put the fractions over a common denominator and the whole function becomes, 12x^13/12 / [8x^3/4 + 3x^1/3 ] both the numerator and denominator still both approach ∞. Using Lhopital's rule derives another function where the numerator and denominator both approach ∞. Looking at the function I think this will always be the case no matter how many times I perform LHopitals rule.

 

[skaa]

You can use this method:
\(\displaystyle =\lim_{x\to \infty}\frac{x^{\frac{2}{3}}(x^{\frac{1}{3}}+x^{-\frac{1}{6}}+x^{-\frac{1}{3}})}{x^{\frac{2}{3}}(1+x^{-\frac{5}{12}})}=\lim_{x\to \infty}\frac{x^{\frac{1}{3}}+x^{-\frac{1}{6}}+x^{-\frac{1}{3}}}{1+x^{-\frac{5}{12}}}=\lim_{x\to \infty}\frac{x^{\frac{1}{3}}}{1}=\infty\)

 

[Ishuda]

[x + x^1/2 + x^1/3] / [x^2/3 + x^1/4 ]

I determine that the numerator and the denominator both approach ∞ so LHopital's rule can be used. when I use lhopital rule I get
[1 + 0.5x^-0.5 +1/(3x^2/3) ] / [2/(3x^1/3) + 1/(4x^3/4) ] . In the numerator, i determine that the term with 1 over a value with x will go to 0 so the numerator becomes just 1, in the denominator i put the fractions over a common denominator and the whole function becomes, 12x^13/12 / [8x^3/4 + 3x^1/3 ] both the numerator and denominator still both approach ∞. Using Lhopital's rule derives another function where the numerator and denominator both approach ∞. Looking at the function I think this will always be the case no matter how many times I perform LHopitals rule.

Expanding on skaa's post, you should have had something like the following: If you have the ratio of two functions, try to simplify the numerator and denominator so that each is a simple common function type times something that has a nice limit. Looking at your particular problem we note that if we factor the numerator into
x + x^1/2 + x^1/3 = x [1 + x^-1/2 + x^-2/3]
where we have factored out the largest power of x. This then behaves like x as x \(\displaystyle \to\infty\).

Now do the same thing with the denominator where 2/3 is the largest exponent
x^2/3 + x^1/4 = x^2/3 [1 + x^-5/12]
which behaves as x^2/3 as x\(\displaystyle \to\infty\)

Now, what can you say about the ratio?