Limits and cube roots

Question

I'm rather stumped at the moment. I can graph the following equation but I'm having trouble solving it algebraically.

$$ \lim_{x\to 1} \frac{\sqrt[3]{x}-1}{\sqrt{x}-1} $$

Where do I start?

Answer 1

Since you are allowed to replace $x$ with $y^6$, $$ \lim_{x\to 1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\lim_{y\to 1}\frac{y^2-1}{y^3-1}=\lim_{y\to 1}\frac{y+1}{y^2+y+1}=\color{red}{\frac{2}{3}}.$$

Answer 2

Multiply through by $\frac{\sqrt{x}+1}{\sqrt{x}+1}$ to get the equivalent limit $$\lim_{x \to 1} \frac{x^{5/6}+x^{1/3}-x^{1/2}-1}{x-1}$$ Then use L'Hospital's rule and plug in $x=1$.

Answer 3

Another way (Taylor expansions, again)*: write $x=1+h$, so that you are looking at the limit when $h\to 0$. Recall that, for $\alpha > 0$ the first-order Taylor expansion of $(1+h)^\alpha$ around $0$ is $$ (1+h)^\alpha = \alpha h + o(h). $$ Then, you can compute your limit as the limit around $0$ of $$ \frac{(1+h)^{1/3}-1}{(1+h)^{1/2}-1} = \frac{1+\frac{1}{3}h+o(h)-1}{1+\frac{1}{2}h+o(h)-1} = \frac{\frac{1}{3}h+o(h)}{\frac{1}{2}h+o(h)} = \frac{2+o(1)}{3+o(1)} \xrightarrow[h\to0]{} \frac{2}{3} $$ * A systematic, if not always most elegant, approach.

Answer 4

With conjugates: the rationale behing conjugates is to use the factorisation of $x^r-1$, so the conjgate expression of $\sqrt[3]{x}-1$ is $\,\sqrt[3]{x^2}+\sqrt[3]{x}+1$, so that : $$\frac{\sqrt[3]{x}-1}{\sqrt x-1}=\frac{\sqrt x+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\xrightarrow[x\to 1]{}\frac23.$$

Answer 5

This solution uses what is called: The conjugate expression. Using \begin{equation*} (a-b)(a+b)=a^{2}-b^{2} \end{equation*} with $a=\sqrt{x}$ and $b=1,$ you get \begin{equation*} \left( \sqrt{x}-1\right) (\sqrt{x}+1)=x-1. \end{equation*} [The conjugate of $(\sqrt{x}-1)$ is $(\sqrt{x}+1)$)]. The same way, using \begin{equation*} (a-b)(a^{2}+ab+b^{2})=a^{3}-b^{3} \end{equation*} with $a=\sqrt[3]{x}$ and $b=1,$ you get \begin{equation*} (\sqrt[3]{x}-1)(\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)=x-1. \end{equation*} [The conjugate of $(\sqrt[3]{x}-1)$ is $(\left( \sqrt[3]{x}\right) ^{2}+\sqrt% [3]{x}+1)$]. Therefore \begin{eqnarray*} \frac{(\sqrt[3]{x}-1)}{\left( \sqrt{x}-1\right) } &=&\frac{(\sqrt[3]{x}-1)}{% \left( \sqrt{x}-1\right) }\frac{(\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}% +1)}{(\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)}\frac{(\sqrt{x}+1)}{(% \sqrt{x}+1)} \\ &=&\frac{(\sqrt[3]{x}-1)\cdot (\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)% }{\left( \sqrt{x}-1\right) \cdot (\sqrt{x}+1)}\cdot \frac{(\sqrt{x}+1)}{% (\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)} \\ &=&\frac{x-1}{x-1}\cdot \frac{(\sqrt{x}+1)}{(\left( \sqrt[3]{x}\right) ^{2}+% \sqrt[3]{x}+1)} \\ &=&\frac{(\sqrt{x}+1)}{(\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)} \end{eqnarray*} and the limit follows \begin{equation*} \lim_{x\rightarrow 1}\frac{(\sqrt[3]{x}-1)}{\left( \sqrt{x}-1\right) }% =\lim_{x\rightarrow 1}\frac{(\sqrt{x}+1)}{(\left( \sqrt[3]{x}\right) ^{2}+% \sqrt[3]{x}+1)}=\frac{(\sqrt{1}+1)}{(\left( \sqrt[3]{1}\right) ^{2}+\sqrt[3]{% 1}+1)}=\frac{2}{3}. \end{equation*}

${\bf NOTE:}$ You can take it as a Definition for further uses: The conjugate expression of $A$ (which may contains radicals) is the expression $B$ (which may contains radicals), if $AB$ contains no radicals! For example, the conjugate of $\sqrt[4]{x}-1$ is obtained from the formula \begin{equation*} (a-b)(a^{3}+a^{2}b+ab^{2}+b^{3})=a^{4}-b^{4} \end{equation*} by taking $a=\sqrt[4]{x}$ and $b=1$.

Answer 6

Here's another variation of the theme:

Since we observe $$\lim_{x\rightarrow 1}{\sqrt[3]{x}-1}=\lim_{x\rightarrow 1}{\sqrt{x}-1}=0$$ we can apply L’Hospital’s Rule

We obtain $$\lim_{x\rightarrow 1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1} =\lim_{x\rightarrow 1}\frac{x^{\frac{1}{3}}-1}{x^{\frac{1}{2}}-1} =\lim_{x\rightarrow 1}\frac{\frac{1}{3}x^{-\frac{2}{3}}}{\frac{1}{2}x^{-\frac{1}{2}}} =\frac{\frac{1}{3}}{\frac{1}{2}} =\frac{2}{3} $$

Note: We apply L’Hospital’s Rule to $\lim_{x\rightarrow 1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}$ in order to cope with the indefinite expression $\frac{0}{0}$. We could ask how the answer by @JackDAurizio deals with it. In fact this is done by cancelling $y-1$ in $\lim_{y\rightarrow 1}\frac{y^3-1}{y^2-1}$. We see this way that $1$ is a removable singularity.