Question

If function is differentiable at a, where a>0 is an element of real numbers, evaluate the following limit

limit f(x)-f(a)/√x-√a as x approaches a​

Answer 1

Answer: We can use algebraic manipulation to simplify the expression:

(f(x) - f(a)) / (sqrt(x) - sqrt(a))

Multiplying the numerator and denominator by the conjugate of the denominator, we get:

[(f(x) - f(a)) / (sqrt(x) - sqrt(a))] * [(sqrt(x) + sqrt(a)) / (sqrt(x) + sqrt(a))]

Simplifying the numerator, we get:

[f(x) - f(a)] * [sqrt(x) + sqrt(a)]

Expanding the numerator, we get:

f(x) * sqrt(x) + f(x) * sqrt(a) - f(a) * sqrt(x) - f(a) * sqrt(a)

We can simplify this expression by subtracting and adding f(a) * sqrt(x) to get:

f(x) * sqrt(x) - f(a) * sqrt(x) + f(a) * sqrt(x) + f(x) * sqrt(a) - f(a) * sqrt(a)

Factoring out common terms in the first two and last two terms, we get:

sqrt(x) * [f(x) - f(a)] + sqrt(a) * [f(x) - f(a)]

Now we can simplify further by factoring out the common factor of [f(x) - f(a)]:

[f(x) - f(a)] * [sqrt(x) + sqrt(a)] / [sqrt(x) - sqrt(a)]

At this point, we can take the limit as x approaches a. Note that this expression is in the indeterminate form 0/0. We can use L'Hopital's rule to evaluate the limit. Differentiating the numerator and denominator with respect to x, we get:

[f'(x) * (sqrt(x) + sqrt(a)) + (f(x) - f(a)) / (2 * sqrt(x))] / [1 / (2 * sqrt(x))]

Taking the limit as x approaches a, we get:

[f'(a) * (sqrt(a) + sqrt(a)) + (f(a) - f(a)) / (2 * sqrt(a))] / [1 / (2 * sqrt(a))]

Simplifying, we get:

2 * f'(a) * sqrt(a)

Therefore, the limit of (f(x) - f(a)) / (sqrt(x) - sqrt(a)) as x approaches a is equal to 2 * f'(a) * sqrt(a).

Explanation: