Math Help - Limit

1. Limit

Hi,

I don't know how to calculate $\lim_{x\to +\infty} \frac{\sqrt{x-1}-\sqrt[6]{x}}{\sqrt[3]{x-1}-\sqrt{x}}$.

2. Originally Posted by bhitroofen01
Hi,

I don't know how to calculate $\lim_{x\to +\infty} \frac{\sqrt{x-1}-\sqrt[6]{x}}{\sqrt[3]{x-1}-\sqrt{x}}$.

try using factorisation.....
use for both numerator n denominator

3. Hello, bhitroofen01!

evaluate: . $\lim_{x\to \infty} \frac{\sqrt{x-1}-\sqrt[6]{x}}{\sqrt[3]{x-1}-\sqrt{x}}$

We have: . $\frac{(x-1)^{\frac{1}{2}} - x^{\frac{1}{6}}} {(x-1)^{\frac{1}{3}} - x^{\frac{1}{2}}}$

$\text{Divide top and bottom by }x^{\frac{1}{2}}:\quad \frac{\dfrac{(x-1)^{\frac{1}{2}}}{x^{\frac{1}{2}}} - \dfrac{x^{\frac{1}{6}}}{x^{\frac{1}{2}}}} {\dfrac{(x-1)^{\frac{1}{3}}}{x^{\frac{1}{2}}} - \dfrac{x^{\frac{1}{2}}}{x^{\frac{1}{2}}}}$

. . . $=\;\frac{\left(\dfrac{x-1}{x}\right)^{\frac{1}{2}} - \dfrac{1}{x^{\frac{1}{3}}}}{ \dfrac{(x-1)^{\frac{1}{3}}}{x^{\frac{1}{6}}\cdot x^{\frac{1}{3}}} - 1}$ . $=\;\; \frac{\left(\dfrac{x-1}{x}\right)^{\frac{1}{2}} - \dfrac{1}{x^{\frac{1}{3}}}} {\dfrac{1}{x^{\frac{1}{6}}}\left(\dfrac{x-1}{x}\right)^{\frac{1}{3}} - 1}$ . $=\;\; \frac{\left(1-\dfrac{1}{x}\right)^{\frac{1}{2}} - \dfrac{1}{x^{\frac{1}{3}}}} {\dfrac{1}{x^{\frac{1}{6}}}\left(1 - \dfrac{1}{x}\right)^{\frac{1}{3}} - 1}$

Therefore: . $\lim_{x\to\infty}\left[ \frac{\left(1-\dfrac{1}{x}\right)^{\frac{1}{2}} - \dfrac{1}{x^{\frac{1}{3}}}} {\dfrac{1}{x^{\frac{1}{6}}}\left(1 - \dfrac{1}{x}\right)^{\frac{1}{3}} - 1} \right]$ . $=\;\;\frac{(1-0)^{\frac{1}{2}} - 0}{0(1-0)^{\frac{1}{3}} - 1} \;\;=\;\;\frac{1}{-1} \;\;=\;\;-1$