A. Factor the numerator as a difference of squares:

c. As

, the contribution of the terms of degree less than 2 becomes negligible, which means we can write

e. Let's first rewrite the root terms with rational exponents:
![\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto1%7D%5Cfrac%7B%5Csqrt%5B3%5Dx-x%7D%7B%5Csqrt%20x-x%7D%3D%5Clim_%7Bx%5Cto1%7D%5Cfrac%7Bx%5E%7B1%2F3%7D-x%7D%7Bx%5E%7B1%2F2%7D-x%7D)
Next we rationalize the numerator and denominator. We do so by recalling


In particular,


so we have

For

and

, we can simplify the first term:

So our limit becomes
Answer:
-24x + 18y
Step-by-step explanation:
6 x -4x = -24x
6 x 3y = 18y
Answer:
A
Step-by-step explanation:
I did it in a on a piece of paper
<em>Answer:</em>
<em>$5 for 8 pens</em>
<em></em>
<em>Step-by-step explanation:</em>
<em>0.625 times 8= 5 dollars</em>
<em>0.625 times 50= 31.25</em>
<em>which is way cheaper</em>