8
$\begingroup$

I have been trying to find the limit, $$\lim _{ x \to 0} \frac{\sqrt[3]{1+6x+3x^2+3x^3+3x^4}-\sqrt[4]{1+8x+4x^2+4x^3-2x^4}}{6x^2}$$

and sort of succeeded. But my $0$ answer doesn't converge with what Wolfram says which is $1/3$. Therefore, I would really appreciate it, if you could give me the right answer and drop a hint on the solution.

  • 1
    As written, this limit goes to infinity, not $1/3$. Double check. – 2017-01-19
  • 1
    As can be seen the numerator is $-2x/3 + o(x)$ so limit does not exist. – 2017-01-19
  • 0
    @YvesDaoust That was a mistake in my edit. – 2017-01-19
  • 1
    ok now it is fine. Use Taylor to get numerator as $2x^{2} + o(x^{2})$. Ignore my previous comment (it was based on earlier version of the question which had a typo). – 2017-01-19
  • 0
    So you've tried and got to the wrong result. Do you expect to learn more by getting the correct solution or by identifying what error you did in your solution? As the question is stated we cannot help you with the later (which I think would be more educational). – 2017-01-19
  • 0
    Mr Singh, can you please specify how to use Taylor in this case? For polynomial expression without the root? I don't see how that would simplify the expression. – 2017-01-19

3 Answers 3

6

Another approach (apart from the one mentioned in my comments) is to use Binomial Theorem. Let the expression be denoted by $$\frac{u - v}{w}$$ and then multiply numerator and denominator by $$t = u^{11} + u^{10}v + \cdots + v^{11}$$ to get $$\frac{u^{12} - v^{12}}{wt}$$ and note that since $u, v$ both tend to $1$ as $x \to 0$ the variable $t \to 12$ and hence the desired limit is $$\frac{1}{72}\lim_{x\to 0}\frac{(1+6x+3x^2+3x^3+3x^4)^{4} - (1+8x+4x^2+4x^3-2x^4)^{3}}{x^{2}}$$ As can be seen the numerator can be written as $$(1 + 24x + 228x^{2} + o(x^{2})) - (1 + 24x + 204x^{2} + o(x^{2}))$$ or $$24x^{2} + o(x^{2})$$ and hence the answer is $24/72 = 1/3$.

On OP's request the solution by Taylor series is provided below.

We have the Taylor series for $(1 + x)^{n}$ for all real $n$ as $$(1 + x)^{n} = 1 + nx + \frac{n(n - 1)}{2!}x^{2} + o(x^{2})$$ as $x\to 0$. Hence $$(1+6x+3x^2+3x^3+3x^4)^{1/3} = 1 + 2x - 3x^{2} + o(x^{2})$$ and $$(1+8x+4x^2+4x^3-2x^4)^{1/4} = 1 + 2x - 5x^{2} + o(x^{2})$$ so the answer now comes easily as $1/3$.

  • 0
    Wow, thank you very much for your work. That really helped! – 2017-01-19
3

Without Taylor nor the generalized Binomial theorem:

Use the identity

$$a^{12}-b^{12}=\\ (a-b)(a^{11}+a^{10}b^{1}+a^{9}b^{2}+a^{8}b^{3}+a^{7}b^{4}+a^{6}b^{5}+a^{5}b^{6}+a^{4}b^{7}+a^{3}b^{8}+a^{2}b^{9}+ab^{10}+b^{11}).$$

By multiplying/dividing by the conjugate "dodecanomial", the numerator becomes the polynomial

$$(1+6x+3x^2+3x^3+3x^4)^4-(1+8x+4x^2+4x^3-2x^4)^3=\\ (1+24x+228x^2+\cdots81x^{16})-(1+24x+204x^2+\cdots-8x^{12}),$$

while the denominator is a sum of $12$ rational expressions that tend to $1$. So after simplification, the limit is

$$\frac{228-204}{6\cdot12}=\frac13.$$

Alternatively, you can pull a factor $(1+2x)$ from both radicals and get

$$\frac{\sqrt[3]{1+\frac{-9x^2-5x^3+3x^4}{(1+2x)^3}}-\sqrt[4]{1+\frac{-20x^2-28x^3-18x^4}{(1+2x)^4}}}{6x^2}.$$

It is not hard to show that the Taylor development will yield

$$\frac{(1-\frac93x^2+\cdots)-(1-\frac{20}4x^2+\cdots)}{6x^2}.$$

  • 1
    Your Taylor approach after taking $(1 + 2x)$ out is really simpler as far as calculation of coefficients is concerned. +1 – 2017-01-19
  • 0
    @ParamanandSingh: yep, "complete the cube" and "complete the quad" :) – 2017-01-19
1

First, do a Binomial expansion of the cube root expressions, simplify the numerator, cancel $x^2$ terms and then conduct the limit. You should get $1/3$.

  • 0
    What do you mean by a binomial expansion of the cube root expressions? None of the polynomials can be factored as $(a+x)^n$ – 2017-01-19