Answer:
C.
Step-by-step explanation:
We want to find the limit:
[tex]\displaystyle \lim_{h\to 0} \frac{\sqrt{2x+2h}-\sqrt{2x}}{h}[/tex]
Let's remove the square roots in the numerator by multiplying both layers by the conjugate. Hence:
[tex]\displaystyle =\lim_{h\to 0} \frac{\sqrt{2x+2h}-\sqrt{2x}}{h}\left(\frac{\sqrt{2x+2h}+\sqrt{2x}}{\sqrt{2x+2h}+\sqrt{2x}}\right)[/tex]
In the numerator, we have the difference of two squares pattern. The difference of two squares is:
[tex](a-b)(a+b)=a^2-b^2[/tex]
So, our numerator is now:
[tex](\sqrt{2x+2h}-\sqrt{2x})(\sqrt{2x+2h}+\sqrt{2x})\\[/tex]
Difference of two squares:
[tex]=(\sqrt{2x+2h})^2-(\sqrt{2x})^2[/tex]
Simplify:
[tex]=(2x+2h)-(2x)[/tex]
In the denominator, we can keep everything the same for now. Thus, our limit is simplified to:
[tex]\displaystyle \lim_{h\to 0} \frac{(2x+2h)-(2x)}{h(\sqrt{2x+2h}+\sqrt{2x})}[/tex]
Subtract:
[tex]\displaystyle \lim_{h\to 0} \frac{(2h)}{h(\sqrt{2x+2h}+\sqrt{2x})}[/tex]
We can cancel the h:
[tex]\displaystyle \lim_{h\to 0} \frac{2}{\sqrt{2x+2h}+\sqrt{2x}}[/tex]
Now we can use direct substitution. So:
[tex]\displaystyle \Rightarrow \frac{2}{\sqrt{2x+2(0)}+\sqrt{2x}}[/tex]
Simplify:
[tex]\displaystyle =\frac{2}{\sqrt{2x}+\sqrt{2x}}[/tex]
Combine like terms:
[tex]\displaystyle =\frac{2}{2\sqrt{2x}}[/tex]
Reduce:
[tex]\displaystyle =\frac{1}{\sqrt{2x}}[/tex]
In conclusion:
[tex]\displaystyle \lim_{h\to 0} \frac{\sqrt{2x+2h}-\sqrt{2x}}{h}=\frac{1}{\sqrt{2x}}[/tex]
Hence, our answer is C.
