Question

Find the limit as h approaches 0

Answer 1

C.

Explanation:

We want to find the limit:

`\displaystyle \lim_(h\to 0) (√(2x+2h)-√(2x))/(h)`

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

`\displaystyle =\lim_(h\to 0) (√(2x+2h)-√(2x))/(h)\left((√(2x+2h)+√(2x))/(√(2x+2h)+√(2x))\right)`

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:

`(√(2x+2h)-√(2x))(√(2x+2h)+√(2x))\\`

Difference of two squares:

`=(√(2x+2h))^2-(√(2x))^2`

Simplify:

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

In the denominator, we can keep everything the same for now. Thus, our limit is simplified to:

`\displaystyle \lim_(h\to 0) ((2x+2h)-(2x))/(h(√(2x+2h)+√(2x)))`

Subtract:

`\displaystyle \lim_(h\to 0) ((2h))/(h(√(2x+2h)+√(2x)))`

We can cancel the h:

`\displaystyle \lim_(h\to 0) (2)/(√(2x+2h)+√(2x))`

Now we can use direct substitution. So:

`\displaystyle \Rightarrow (2)/(√(2x+2(0))+√(2x))`

Simplify:

`\displaystyle =(2)/(√(2x)+√(2x))`

Combine like terms:

`\displaystyle =(2)/(2√(2x))`

Reduce:

`\displaystyle =(1)/(√(2x))`

In conclusion:

`\displaystyle \lim_(h\to 0) (√(2x+2h)-√(2x))/(h)=(1)/(√(2x))`

Hence, our answer is C.