1. ## Hard Limit

How do get the following

$\displaystyle\lim_{x\to 1}\frac{x^\frac{1}{7}-1}{x^\frac{1}{5}-1}$

Without l'Hopital's Rule....

2. One way to appraoch it is to do some fancy factoring.

$\lim_{x\rightarrow{1}}\frac{x^{\frac{1}{7}}-1}{x^{\frac{1}{5}}-1}$

= $\lim_{x\rightarrow{1}}\frac{x^{\frac{4}{35}}+x^{\f rac{3}{35}}+x^{\frac{2}{35}}+x^{\frac{1}{35}}+1}{x ^{\frac{6}{35}}+x^{\frac{4}{35}}+x^{\frac{3}{35}}+ x^{\frac{2}{35}}+x^{\frac{1}{7}}+x^{\frac{1}{35}}+ 1}=\boxed{\frac{5}{7}}$

3. Originally Posted by polymerase
$\displaystyle\lim_{x\to 1}\frac{x^\frac{1}{7}-1}{x^\frac{1}{5}-1}$
Substitute $u^{35}=x,$ the limit becomes to

$\lim_{u\to1}\frac{u^5-1}{u^7-1}.$

The factor $u-1$ is bothering the top & bottom, so we pull it out:

$u^5-1=(u-1)(u^4+u^3+u^2+u+1),$ $u^7-1=(u-1)(u^6+u^5+u^4+u^3+u^2+u+1),$ so

$\lim_{u\to1}\frac{u^5-1}{u^7-1}=\lim_{u\to1}\frac{u^4+u^3+u^2+u+1}{u^6+u^5+u^4+ u^3+u^2+u+1},$

and the conclusion follows $\blacksquare$

4. Originally Posted by galactus
One way to appraoch it is to do some fancy factoring.

$\lim_{x\rightarrow{1}}\frac{x^{\frac{1}{7}}-1}{x^{\frac{1}{5}}-1}$

= $\lim_{x\rightarrow{1}}\frac{x^{\frac{4}{35}}+x^{\f rac{3}{35}}+x^{\frac{2}{35}}+x^{\frac{1}{35}}+1}{x ^{\frac{6}{35}}+x^{\frac{4}{35}}+x^{\frac{3}{35}}+ x^{\frac{2}{35}}+x^{\frac{1}{7}}+x^{\frac{1}{35}}+ 1}=\boxed{\frac{5}{7}}$
How do you do that....what's the "rule"(lack of a better word)?

5. It's basically the same as the previous post, but K eliminated the fractional exponents. Just factoring. See the pattern?.

6. Originally Posted by galactus
It's basically the same as the previous post, but K eliminated the fractional exponents. Just factoring. See the pattern?.
I get K's completely but i dont see the pattern for yours

7. Originally Posted by polymerase
I get K's completely but i dont see the pattern for yours
Its just writting $x^{1/7}-1$ and $x^{1/5}-1$ as sums of powers of $x^{1/35}$ , which can be done because $5\times 7=35$

RonL