It has to be true because
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:
46784
Step-by-step explanation:
You would multiply all.the numbers and then your w
answer will be 46784
In mathematics, the additive inverse of a number a is the number that, when added to a yields zero. This operation is also known as the opposite (number), sign change, and negation. ... For example, the additive inverse of 7 is −7, because 7 + (−7) = 0, and the additive inverse of −0.3 is 0.3, because −0.3 + 0.3 = 0 .