Solving indeterminate limit form without L'Hopital's rule

Question

$$\lim_{x\to2}\frac{\sqrt[4]{x}-\sqrt[4]{2}}{\sqrt[3]{x}-\sqrt[3]{2}}$$

Using L'Hopital's rule the simplified form of the above limit is found to be $\frac{3}{4\sqrt[12]{2}}$

However, is it possible to simplify it without using derivatives? Trying to rationalize the denominator using difference of cubes leads to:

Answer 1

Let $a=\sqrt[12]{x}$ and $b=\sqrt[12]{2}$.

$$\lim_{x\to2}\frac{\sqrt[4]{x}-\sqrt[4]{2}}{\sqrt[3]{x}-\sqrt[3]{2}}=\lim_{a\to b}\frac{a^3-b^3}{a^4-b^4}=\lim_{a\to b}\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a+b)(a^2+b^2)}=\lim_{a\to b}\frac{a^2+ab+b^2}{(a+b)(a^2+b^2)}.$$ Can you take it from here?

Answer 2

So you have $$L=\lim_{x\to2}\frac{x^{\frac{1}{4}}-2^{\frac{1}{4}}}{x^{\frac{1}{3}}-2^{\frac{1}{3}}}$$ Note $\text{lcm}(3,4)=12$ thus let $x=u^{12}$ so that as $x\to2 $ we'll have $u\to\sqrt[12]{2}$. Thus $$L=\lim_{u\to\sqrt[12]{2}}\frac{(u^{12})^{\frac{1}{4}}-\sqrt[4]{2}}{(u^{12})^{\frac{1}{3}}-\sqrt[3]{2}}=\lim_{u\to\sqrt[12]{2}}\frac{u^3-c}{u^4-d}$$

where $c=\sqrt[4]{2}$ and $d=\sqrt[3]{2}$. Factoring $$L=\lim_{u\to\sqrt[12]{2}}\frac{(u-\sqrt[3]{c})(u^2+u\sqrt[3]{c}+\sqrt[3]{c^2})}{(u-\sqrt{\sqrt{d}})(u+\sqrt{\sqrt{d}})(u^2+\sqrt{d})}=\lim_{u\to\sqrt[12]{2}}\frac{(u-\sqrt[12]{2})(u^2+\sqrt[12]{2}u+\sqrt[6]{2})}{(u-\sqrt[12]{2})(u+\sqrt[12]{2})(u^2+\sqrt[6]{2})}$$ $$=\lim_{u\to\sqrt[12]{2}}\frac{u^2+\sqrt[12]{2}u+\sqrt[6]{2}}{(u+\sqrt[12]{2})(u^2+\sqrt[6]{2})}=\frac{(2^{\frac{1}{12}})^2+(2^{\frac{1}{12}})^2+\sqrt[6]{2}}{2\sqrt[12]{2}\cdot((2^{\frac{1}{12}})^2+\sqrt[6]{2})}$$ $$=\frac{3\sqrt[6]{2}}{4\sqrt[4]{2}}=\frac{3}{4\sqrt[12]{2}}$$

Answer 3

$\sqrt[4]{x}\ -\ \sqrt[4]{2}\ =\ (\sqrt[12]{x}\ -\ \sqrt[12]{2})(\sqrt[12]{x^2}\ +\ \sqrt[12]{2x}\ +\ \sqrt[12]{2^2})$

$\sqrt[3]{x}\ -\ \sqrt[3]{2}\ =\ (\sqrt[12]{x}\ +\ \sqrt[12]{2})(\sqrt[12]{x^3}\ +\ \sqrt[12]{2x^2}\ +\ \sqrt[12]{4x}\ +\sqrt[12]{8})$

Now cancelling the common factor and putting your limit you will get your required answer.

Answer 4

Start with $x=y+2$ to make $$\lim_{x\to2}\frac{\sqrt[4]{x}-\sqrt[4]{2}}{\sqrt[3]{x}-\sqrt[3]{2}}=\lim_{y\to0}\frac{\sqrt[4]{2+y}-\sqrt[4]{2}}{\sqrt[3]{2+y}-\sqrt[3]{2}}$$ and use the binomial expansion $$\sqrt[4]{2+y}=\sqrt[4]{2}+\frac{y}{4\ 2^{3/4}}-\frac{3 y^2}{64\ 2^{3/4}}+\cdots$$ $$\sqrt[3]{2+y}=\sqrt[3]{2}+\frac{y}{3\ 2^{2/3}}-\frac{y^2}{18\ 2^{2/3}}+\cdots$$ $$\frac{\sqrt[4]{2+y}-\sqrt[4]{2}}{\sqrt[3]{2+y}-\sqrt[3]{2}}=\frac{\frac{y}{4\ 2^{3/4}}-\frac{3 y^2}{64\ 2^{3/4}} +\cdots}{\frac{y}{3\ 2^{2/3}}-\frac{y^2}{18\ 2^{2/3}}+\cdots }$$ Now, long division to get $$\frac{\sqrt[4]{2+y}-\sqrt[4]{2}}{\sqrt[3]{2+y}-\sqrt[3]{2}}=\frac{3}{4 \sqrt[12]{2}}-\frac{y}{64 \sqrt[12]{2}}+\cdots=\frac{3}{4 \sqrt[12]{2}}-\frac{x-2}{64 \sqrt[12]{2}}+\cdots$$ which shows the limit and how it is approached.

Edit

Doing the same for $$\lim_{x\to a}\frac{\sqrt[m]{x}-\sqrt[m]{a}}{\sqrt[n]{x}-\sqrt[n]{a}}$$ you could establish that $$\frac{\sqrt[m]{x}-\sqrt[m]{a}}{\sqrt[n]{x}-\sqrt[n]{a}}=\frac{n }{m}a^{\frac{1}{m}-\frac{1}{n}}+\frac{(n-m) }{2 m^2}a^{\frac{1}{m}-\frac{1}{n}-1}(x-a)+\cdots$$ that is to say $$\frac{n }{m}a^{\frac{n-m}{m n}}\left(1+\frac{n-m}{2 amn}(x-a) \right)+\cdots$$