Derivative of 2sqrt(x) using limits

To find the derivative of the function f(x) = 2√x using limits, we can apply the definition of the derivative. The derivative of a function f(x) at a point x = a is defined as the limit of the difference quotient as x approaches a.

Let's start by finding the difference quotient:

f'(x) = lim(h->0) [f(x+h) - f(x)] / h

In our case, f(x) = 2√x, so let's substitute it into the above formula:

f'(x) = lim(h->0) [2√(x+h) - 2√x] / h

Now we need to simplify the expression inside the limit:

f'(x) = lim(h->0) [2√(x+h) - 2√x] / h
= lim(h->0) [2√(x+h) - 2√x] / h * [√(x+h) + √x] / [√(x+h) + √x]

To simplify further, we can multiply the numerator and the denominator by [√(x+h) - √x]:

f'(x) = lim(h->0) [2√(x+h) - 2√x] / h * [√(x+h) + √x] / [√(x+h) + √x] * [√(x+h) - √x] / [√(x+h) - √x]
= lim(h->0) [2(x+h) - 2x] / h * [√(x+h) - √x] / [√(x+h) + √x]

Now we can cancel out the terms in the numerator:

f'(x) = lim(h->0) [2h] / h * [√(x+h) - √x] / [√(x+h) + √x]
= lim(h->0) 2 * [√(x+h) - √x] / [√(x+h) + √x]

Finally, we can simplify the expression inside the limit by multiplying the numerator and the denominator by the conjugate of the denominator:

f'(x) = lim(h->0) 2 * [√(x+h) - √x] / [√(x+h) + √x] * [√(x+h) - √x] / [√(x+h) - √x]
= lim(h->0) 2 * [(√(x+h))^2 - (√x)^2] / [(√(x+h) + √x) * (√(x+h) - √x)]
= lim(h->0) 2 * [x + h - x] / [(√(x+h) + √x) * (√(x+h) - √x)]
= lim(h->0) 2h / [(√(x+h) + √x) * (√(x+h) - √x)]

Now, as h approaches 0, the fraction in the denominator [√(x+h) + √x] * (√(x+h) - √x) approaches 0 since it contains the term h. Therefore, we can substitute h = 0 into the numerator as well:

f'(x) = 2 * [0] / [0 * (√(x) - √x)]
= 0

Hence, the derivative of the function f(x) = 2√x using limits is 0.