$\dfrac{\sqrt{x+4}-2}x\cdot\dfrac{\sqrt{x+4}+2}{\sqrt{x+4}+2}=\dfrac{(\sqrt{x+4})^2-2^2}{x(\sqrt{x+4}+2)}=\dfrac x{x(\sqrt{x+4}+2)}=\dfrac1{\sqrt{x+4}+2}$

This is continuous at $x=0$ , so we can evaluate the limit directly by substitution:

$\displaystyle\lim_{x\to0}\frac{\sqrt{x+4}-2}x=\lim_{x\to0}\frac1{\sqrt{x+4}+2}=\frac1{\sqrt4+2}=\frac14$

5 0

3 years ago

without building a model tell whether a long row of 8 cubes or a cube made from 8 would have a greater volume. explain

Viktor [21]

I'm presuming that the 'large cube' is made from 8 cubes the same size as the 8 cubes in the long row. They both have the same volume, the only difference between them is that they are arranged in a different order. Hope this helps. If not feel free to ask again and give more information.

7 0

3 years ago