Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider this limit:

$$ \lim_{x \to 1} \frac{\sqrt x - 1}{ \sqrt[3] x - 1} $$ The answer is given to be 2 in the textbook. Our math professor skipped this question telling us it is not in our syllabus, but how can it be solved?

share|cite|improve this question
7  
Surely the answer is $\frac 32$, not $2$. Easy enough to compute values near $x=1$ to see that $2$ isn't right. To prove it...have you seen L'Hopital's rule? – lulu Jan 24 at 11:08
    
I have no idea abour hospital's rule :/ – Ammaar Solkar Jan 24 at 11:09
3  
Well, I expect that's why it was skipped over. It's not that complicated (assuming you know derivatives)...you could look it up. A (more or less equivalent) way is to use a generalized binomial theorem to write $\sqrt{1+\epsilon}\sim 1+\frac {\epsilon}2$, $\sqrt[3]{1+\epsilon}\sim 1+\frac {\epsilon}3$ but again you probably need some calculus. – lulu Jan 24 at 11:15
    
This is basically the same question as here and here and here. – Martin Sleziak Jan 24 at 20:07
    
A general advice: Before asking here try to search whether the same question was asked here before. Or the very least you can do is to check list of similar questions which the software generates when you are posting the questions. And also the list of similar questions which appears in the sidebar after you post the question. If you include the limit in the title, it increases the chance that the questions the software finds have similar titles and are about similar problemss. – Martin Sleziak Jan 24 at 20:34
up vote 17 down vote accepted

Here's a similar approach to Spencer and Harish, but I use a substitution to make it a little easier to read.

First, we eliminate the fractional exponents by substituting $x=u^6$.
Note that $\lim_{x \to 1} u = x = 1$

$$\begin{align}\\ & \lim_{x \to 1} \frac{\sqrt x - 1}{ \sqrt[3] x - 1}\\ = & \lim_{u \to 1} \frac{u^3 - 1}{u^2 - 1}\\ = & \lim_{u \to 1} \frac{(u - 1)(u^2+u+1)}{(u-1)(u+1)}\\ = & \lim_{u \to 1} \frac{u^2+u+1}{u+1}\\ = & \frac{3}{2} \end{align}$$

share|cite|improve this answer

Note the following identities,

$$ y^2-1 = (y-1)(y+1) $$

$$ y^3-1 = (y-1)(y^2+y+1) $$

we can use these to rewrite the numerator and the denomiantor by substituting $\sqrt{x}$ and $\sqrt[3]{x}$ for $y$ respectively,

$$ x - 1 = (\sqrt{x}-1)(\sqrt{x}+1) \Rightarrow \sqrt{x}-1 = \frac{x-1}{\sqrt{x}+1} $$

$$ x - 1 = (\sqrt[3]{x}-1)(\sqrt[3]{x}^2+\sqrt[3]{x}+1) \Rightarrow \sqrt[3]{x}-1 = \frac{x-1}{\sqrt[3]{x}^2+\sqrt[3]{x}+1} $$

substituting this into the ratio you're taking the limit of we get,

$$ \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{\sqrt[3]{x}-1} = \lim_{x\rightarrow 1} \frac{x-1}{\sqrt{x}+1} \frac{\sqrt[3]{x}^2+\sqrt[3]{x}+1} {x-1} = \frac32 $$

share|cite|improve this answer

Using L'Hopital, $$\lim_{x\rightarrow 1} \frac{x^{1/2}-1}{x^{1/3}{-1}} = \lim_{x\rightarrow 1} \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{3x^{2/3}}} = \frac{3}{2}$$

share|cite|improve this answer

The limit is solved as follows:

$$\lim_{x \to 1} \frac{\sqrt x - 1}{ \sqrt[3] x - 1} $$ $$=\lim_{x \to 1} \frac{\frac{\sqrt x - 1}{x-1}}{ \frac{\sqrt[3] x - 1}{x-1}} $$ $$=\frac{\lim_{x \to 1} \frac{\sqrt x - 1}{x-1}}{ \lim_{x \to 1} \frac {\sqrt[3] x - 1}{x-1}} $$ $$=\frac{\frac{1}{2}\cdot 1^{-\frac{1}{2}}}{\frac{1}{3}\cdot 1^{-\frac{2}{3}}} $$ $$=\frac{3}{2}$$

using the common limit formula : $$\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$$

share|cite|improve this answer

Notice, $$\lim_{x\to 1}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}$$ $$=\lim_{x\to 1}\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{(\sqrt[3]{x}-1)(x^{2/3}+1+\sqrt [3]x)}\cdot \frac{(x^{2/3}+1+\sqrt [3]x)}{(\sqrt x+1)}$$ $$=\lim_{x\to 1}\frac{(x-1)}{(x-1)}\cdot \frac{(x^{2/3}+1+\sqrt[3] x)}{(\sqrt x+1)}$$ $$=\lim_{x\to 1}\frac{(x^{2/3}+1+\sqrt[3] x)}{(\sqrt x+1)}$$ $$=\frac{1+1+1}{1+1}=\color{red}{\frac 32}$$

share|cite|improve this answer

As lcm$(2,3)=6$

let $\sqrt[6]x=y\implies\sqrt[3]x=y^2, \sqrt x=y^3$

$$\lim_{x\to1}\dfrac{\sqrt x-1}{\sqrt[3]x-1}=\lim_{y\to1}\dfrac{y^3-1}{y^2-1}=\lim_{y\to1}\dfrac{(y-1)(y^2+y+1)}{(y-1)(y+1)}$$

Safely cancel out $y-1$ as $y\to1, y-1\to0\implies y-1\ne0$

OR

set $\sqrt[6]x=y+1$ to get $$\lim_{x\to1}\dfrac{\sqrt x-1}{\sqrt[3]x-1}=\lim_{y\to0}\dfrac{(1+y)^3-1}{(1+y)^2-1}=\lim_{y\to0}\dfrac{3y+3y^2+y^3}{2y+y^2}=\cdots$$

share|cite|improve this answer

Another way : change variable $x=1+y$; so $$A=\frac{\sqrt x - 1}{ \sqrt[3] x - 1}=\frac{\sqrt{1+y} - 1}{ \sqrt[3] {1+y} - 1}$$ Now, using the fact that, close to $y=0$ (using the generalized binomial theorem as lulu commented) $$(1+y)^a=1+a y+\frac{1}{2}a \left(a-1\right) y^2+O\left(y^3\right)$$ which makes $$A=\frac{1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)-1}{1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)-1}\approx \frac{\frac{y}{2}-\frac{y^2}{8}}{\frac{y}{3}-\frac{y^2}{9}}=\frac{\frac{1}{2}-\frac{y}{8}}{\frac{1}{3}-\frac{y}{9}}$$ Now make $y\to0$ to get the result.

You can even get more if you know long division. Omitting the high order terms, the last expression is $\sim\frac{3}{2}+\frac{y}{8}$ which reveals not only the limit but also how it is approached.

share|cite|improve this answer
    
Good manipulation. But I have seen a lot of limit of the expression problems but I did not understand what &O(y^3)& means and how is it eliminated from the numerator and the denominator. – Vinay5forPrime Jan 24 at 16:01
    
This is the Big O notation. Don't worry, you will learn it very soon. It describes the limiting behavior of a function when the argument tends towards a particular value. My answer was just to show another approach (you will also learn it soon). Cheers. – Claude Leibovici Jan 24 at 16:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.