Question

Find the limit as h approaches 0

Answer 1

Answer:

C.

Step-by-step explanation:

So we want to find:

$\lim_{h\to 0} \frac{\sqrt{2x+2h}-\sqrt{2x}}{h}$

Let's remove the square roots in the numerator by multiplying both layers by the conjugate. So:

$=\lim_{h\to 0} \frac{\sqrt{2x+2h}-\sqrt{2x}}{h}(\frac{\sqrt{2x+2h}+\sqrt{2x}}{\sqrt{2x+2h}+\sqrt{2x}})$

In the numerator, we have the difference of two squares pattern. The difference of two squares is:

$(a-b)(a+b)=a^2-b^2$

So, our numerator is now:

$(\sqrt{2x+2h}-\sqrt{2x})(\sqrt{2x+2h}+\sqrt{2x})$

Difference of two squares:

$=(\sqrt{2x+2h})^2-(\sqrt{2x})^2$

Simplify:

$=(2x+2h)-(2x)$

In the denominator, keep everything the same for now. So, our limit is now:

$\lim_{h\to 0} \frac{(2x+2h)-(2x)}{h(\sqrt{2x+2h}+\sqrt{2x})}$

Subtract in the numerator:

$\lim_{h\to 0} \frac{(2h)}{h(\sqrt{2x+2h}+\sqrt{2x})}$

Both the numerator and denominator have a h. Cancel:

$\lim_{h\to 0} \frac{(2)}{(\sqrt{2x+2h}+\sqrt{2x})}$

Now, we can use direct substitution. Substitute 0 for h. So:

$=\frac{2}{(\sqrt{2x+2(0)}+\sqrt{2x})}$

Simplify:

$=\frac{2}{(\sqrt{2x}+\sqrt{2x})}$

Combine like terms:

$=\frac{2}{2\sqrt{2x}}$

Reduce:

$=\frac{1}{\sqrt{2x}}$

Therefore, our limit is equal to:

$\lim_{h\to 0} \frac{\sqrt{2x+2h}-\sqrt{2x}}{h}=\frac{1}{\sqrt{2x}}$

So, our answer is C.