Answer:
C.
Step-by-step explanation:
So we want to find:
![\lim_{h\to 0} \frac{\sqrt{2x+2h}-\sqrt{2x}}{h}](https://tex.z-dn.net/?f=%5Clim_%7Bh%5Cto%200%7D%20%5Cfrac%7B%5Csqrt%7B2x%2B2h%7D-%5Csqrt%7B2x%7D%7D%7Bh%7D)
Let's remove the square roots in the numerator by multiplying both layers by the conjugate. So:
![=\lim_{h\to 0} \frac{\sqrt{2x+2h}-\sqrt{2x}}{h}(\frac{\sqrt{2x+2h}+\sqrt{2x}}{\sqrt{2x+2h}+\sqrt{2x}})](https://tex.z-dn.net/?f=%3D%5Clim_%7Bh%5Cto%200%7D%20%5Cfrac%7B%5Csqrt%7B2x%2B2h%7D-%5Csqrt%7B2x%7D%7D%7Bh%7D%28%5Cfrac%7B%5Csqrt%7B2x%2B2h%7D%2B%5Csqrt%7B2x%7D%7D%7B%5Csqrt%7B2x%2B2h%7D%2B%5Csqrt%7B2x%7D%7D%29)
In the numerator, we have the difference of two squares pattern. The difference of two squares is:
![(a-b)(a+b)=a^2-b^2](https://tex.z-dn.net/?f=%28a-b%29%28a%2Bb%29%3Da%5E2-b%5E2)
So, our numerator is now:
![(\sqrt{2x+2h}-\sqrt{2x})(\sqrt{2x+2h}+\sqrt{2x})\\](https://tex.z-dn.net/?f=%28%5Csqrt%7B2x%2B2h%7D-%5Csqrt%7B2x%7D%29%28%5Csqrt%7B2x%2B2h%7D%2B%5Csqrt%7B2x%7D%29%5C%5C)
Difference of two squares:
![=(\sqrt{2x+2h})^2-(\sqrt{2x})^2](https://tex.z-dn.net/?f=%3D%28%5Csqrt%7B2x%2B2h%7D%29%5E2-%28%5Csqrt%7B2x%7D%29%5E2)
Simplify:
![=(2x+2h)-(2x)](https://tex.z-dn.net/?f=%3D%282x%2B2h%29-%282x%29)
In the denominator, keep everything the same for now. So, our limit is now:
![\lim_{h\to 0} \frac{(2x+2h)-(2x)}{h(\sqrt{2x+2h}+\sqrt{2x})}](https://tex.z-dn.net/?f=%5Clim_%7Bh%5Cto%200%7D%20%5Cfrac%7B%282x%2B2h%29-%282x%29%7D%7Bh%28%5Csqrt%7B2x%2B2h%7D%2B%5Csqrt%7B2x%7D%29%7D)
Subtract in the numerator:
![\lim_{h\to 0} \frac{(2h)}{h(\sqrt{2x+2h}+\sqrt{2x})}](https://tex.z-dn.net/?f=%5Clim_%7Bh%5Cto%200%7D%20%5Cfrac%7B%282h%29%7D%7Bh%28%5Csqrt%7B2x%2B2h%7D%2B%5Csqrt%7B2x%7D%29%7D)
Both the numerator and denominator have a h. Cancel:
![\lim_{h\to 0} \frac{(2)}{(\sqrt{2x+2h}+\sqrt{2x})}](https://tex.z-dn.net/?f=%5Clim_%7Bh%5Cto%200%7D%20%5Cfrac%7B%282%29%7D%7B%28%5Csqrt%7B2x%2B2h%7D%2B%5Csqrt%7B2x%7D%29%7D)
Now, we can use direct substitution. Substitute 0 for h. So:
![=\frac{2}{(\sqrt{2x+2(0)}+\sqrt{2x})}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2%7D%7B%28%5Csqrt%7B2x%2B2%280%29%7D%2B%5Csqrt%7B2x%7D%29%7D)
Simplify:
![=\frac{2}{(\sqrt{2x}+\sqrt{2x})}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2%7D%7B%28%5Csqrt%7B2x%7D%2B%5Csqrt%7B2x%7D%29%7D)
Combine like terms:
![=\frac{2}{2\sqrt{2x}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2%7D%7B2%5Csqrt%7B2x%7D%7D)
Reduce:
![=\frac{1}{\sqrt{2x}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2x%7D%7D)
Therefore, our limit is equal to:
![\lim_{h\to 0} \frac{\sqrt{2x+2h}-\sqrt{2x}}{h}=\frac{1}{\sqrt{2x}}](https://tex.z-dn.net/?f=%5Clim_%7Bh%5Cto%200%7D%20%5Cfrac%7B%5Csqrt%7B2x%2B2h%7D-%5Csqrt%7B2x%7D%7D%7Bh%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2x%7D%7D)
So, our answer is C.