0

I know the correct answer is $\frac{16a}{9}$ but when I substitute directly I get $2a$ and I see no reason why I shouldn't substitute directly since it's straight forward.

Km356
  • 842
  • 1
    How do you get $2a$? – Botond May 12 '19 at 20:28
  • 2
    you don't get $2a$, you get an undefined result, as the denominator becomes zero when you plug in. – The Count May 12 '19 at 20:30
  • 2
    @Botond I think this is done as $\frac {a^2-a^2}{a-a}=a+a=2a$ which is ingenious, but not correct, because the limit requires consideration of the first order terms as well as the zeroth order ones. – Mark Bennet May 12 '19 at 20:30
  • Looks like if you plug $x=a$ into the denominator (and $a>0$), you get $a-\sqrt[4]{aa^3}=a-\sqrt[4]{a^4} = a-a = 0.$ – Mankind May 12 '19 at 20:31
  • please explain what you mean by consideration of the zeroth order terms @MarkBennet – Km356 May 12 '19 at 20:33
  • @Km356 Did you really just say that $$\frac00=\frac{0\times k}0=k$$? Because that is not a valid statement. – Peter Foreman May 12 '19 at 20:39
  • I'd say the're not really zero and that they just approach it and therefore are still numbers that can be cancelled out – Km356 May 12 '19 at 20:42
  • 2
    Suppose $x=a+d$. The "order" of the terms in this case is determined by the power of the (assumed small) difference $d$. The zeroth order terms cancel top and bottom (which is why you get the indeterminate form $\frac 00$ when you substitute $x=a$) which mean the terms of first order in $d$ will be the ones to determine the value (provided they don't both vanish too). Since they are all of first order the $d$'s will cancel. The first order terms are isolated by using binomial expansion, or by using L'Hopital's rule (which uses derivatives to the same effect and works more generally). – Mark Bennet May 12 '19 at 20:44
  • @Km356, your approach would be correct if you actually get $\frac{x^2-a^2}{x-a}$ so you can factor $x-a$(which is not 0) and cancel it. But this is not the case, your function is different , more complicated than this. Just to reinforce this, $\frac{ax-x^2}{a-x}$ has the form $\frac{a^2-a^2}{a-a}$, but guess what, when you take the limit as $x\to a$, you get $a$, not $2a$. – Julian Mejia May 12 '19 at 21:24
  • You wrote "I'd say the're not really zero and that they just approach it and therefore are still numbers that can be cancelled out", but that's a misunderstanding of the substitution method. The substituion method requires that when you substitute $x=a$, the result is a valid expression of a real number. The fraction $\frac{a^2-a^2}{a-a}$ is not a valid expression of a real number, no matter what the value of $a$, as the comment of @PeterForeman shows. – Lee Mosher May 13 '19 at 02:59

3 Answers3

3

Simplifying first, then letting $a/x = 1+c$, so $x = a(1-c+O(c^2))$ as $c \to 0$,

$\begin{array}\\ D &=\dfrac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}\\ &=\dfrac{x^2\sqrt{2a^3/x^3-1}-ax\sqrt[3]{a^2/x^2}}{a-x\sqrt[4]{a/x}}\\ &=\dfrac{x^2\sqrt{2(1+c)^3-1}-ax\sqrt[3]{(1+c)^2}}{x(a/x-\sqrt[4]{1+c})}\\ &=\dfrac{x\sqrt{2(1+3c+3c^2+c^3)-1}-a(1+c)^{2/3}}{1+c-\sqrt[4]{1+c}}\\ &=\dfrac{x\sqrt{1+6c+O(c^2)}-a(1+2c/3+O(c^2))}{1+c-(1+c/4+O(c^2)}\\ &=\dfrac{a(1-c+O(c^2))(1+3c+O(c^2))-a(1+2c/3+O(c^2))}{3c/4+O(c^2)}\\ &=\dfrac{a(1+2c+O(c^2))-a(1+2c/3+O(c^2))}{3c/4+O(c^2)}\\ &=\dfrac{(4/3)ac+O(c^2))}{3c/4+O(c^2)}\\ &=(16/9)a+O(c)\\ \end{array} $

marty cohen
  • 107,799
1

Limit of a function at a point is not the same as the value of the function at that point so substituting $x=a$ is not the right approach. See more details in this answer.

Instead use the theorems related to evaluation of limits. Since the function involved here is algebraic the following limit formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$ can be used. The denominator of the expression in question can be written as $$-a^{1/4}\cdot\frac{x^{3/4}-a^{3/4}}{x-a}\cdot(x-a)$$ and the fraction in middle tends to $(3/4)a^{-1/4}$ so that the denominator can be replaced by $$-\frac{3}{4}\cdot(x-a)$$ The numerator in the expression in question is slightly complicated and one can split it into two parts to get $$\{\sqrt{2a^3x-x^4}-a^2\} - \{a\sqrt[3]{a^2x}-a^2\}$$ For the first part let $t=2a^3x-x^4$ so that $t\to a^4$ as $x\to a$. We have $$\frac{\sqrt{t} - a^2}{x-a}=\frac{t^{1/2}-(a^4)^{1/2}}{t-a^4}\cdot\frac{t-a^4}{x-a}$$ The first fraction tends to $(1/2)a^{-2}$ and $$\frac{t-a^4}{x-a}=\frac{2a^3x-2a^4}{x-a}-\frac{x^4-a^4}{x-a}$$ which tends to $2a^3-4a^3=-2a^3$. It follows that $$\frac{\sqrt{t} - a^2}{(-3/4)(x-a)}\to \frac{4a}{3}$$ Similarly $$\frac{a\sqrt[3]{a^2x}-a^2}{-(3/4)(x-a)}=-\frac{4}{3}\cdot a^{5/3}\cdot\frac{x^{1/3}-a^{1/3}}{x-a}\to - \frac{4}{3}\cdot a^{5/3}\cdot\frac{1}{3}\cdot a^{-2/3}=-\frac{4a}{9}$$ The desired limit is thus $(4a/3)+(4a/9)=16a/9$.

Paramanand Singh
  • 87,309
1

When you go from $\frac{a^2-a^2}{a-a}$ to $\frac{a+a}{1}$, you are dividing numerator and denominator by $a-a=0$. $$ \begin{align} \lim_{x\to a}\frac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}} &=\lim_{x\to1}\frac{a^2\sqrt{2x-x^4}-a^2\sqrt[3]{\vphantom{2}x}}{a-a\sqrt[4]{x^3}}\tag1\\ &=a\lim_{x\to1}\frac{\sqrt{2x-x^4}-\sqrt[3]{\vphantom{2}x}}{1-\sqrt[4]{x^3}}\tag2\\ &=a\lim_{x\to0}\frac{\sqrt{1-2x-6x^2-4x^3-x^4}-\sqrt[3]{1+x}}{1-\sqrt[4]{1+3x+3x^2+x^3}}\tag3\\ &=a\lim_{x\to0}\frac{1-x+O\!\left(x^2\right)-1-\frac13x+O\!\left(x^2\right)}{1-1-\frac34x+O\!\left(x^2\right)}\tag4\\ &=a\lim_{x\to0}\frac{-\frac43x+O\!\left(x^2\right)}{-\frac34x+O\!\left(x^2\right)}\tag5\\ &=a\lim_{x\to0}\frac{-\frac43+O(x)}{-\frac34+O(x)}\tag6\\ &=a\lim_{x\to0}\left(\frac{16}9+O(x)\right)\tag7\\[6pt] &=\frac{16}9a\tag8 \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto ax$
$(2)$: pull the factors of $a$ out front
$(3)$: substitute $x\mapsto1+x$
$(4)$: apply Taylor series
$(5)$: simplify numerator and denominator
$(6)$: divide numerator and denominator by $x$
$(7)$: divide
$(8)$: evaluate the limit

robjohn
  • 345,667