find the limit of f'(x) = 1/(√x) using the limit definition of derivative as x approaches 0.
Given f(x)=1/√x
f'(x)
=Lim h->0 (f(x+h)-f(x))/h
=Lim h->0 (1/√(x+h)-1/√x))/h
subtract with common denominator
=Lim h->0 ((√x-√(x+h)/(h(√x √(x+h)))
multiply by conjugate of numerator, √(x)+√(x+h)
=Lim h->0 (x-(x+h))/(h(√x √(x+h))*(√x+&radic(x+h)))
subtract and cancel h
=Lim h->0 -1/(√x √(x+h)*(√x+&radic(x+h)))
Take limit h->0
=-1/x3/2
To find the limit of the derivative f'(x) = 1/(√x) as x approaches 0 using the limit definition of derivative, we need to follow these steps:
Step 1: Write down the limit definition of the derivative. The limit definition of the derivative is given by:
f'(x) = lim(h->0) [f(x + h) - f(x)] / h
Step 2: Replace f(x) with 1/(√x) in the limit definition:
f'(x) = lim(h->0) [1/(√(x + h)) - 1/(√x)] / h
Step 3: Simplify the expression inside the limit:
f'(x) = lim(h->0) [(√x - √(x + h)) / (√x * √(x + h))] / h
Step 4: To simplify the numerator, we multiply by the conjugate:
f'(x) = lim(h->0) [(√x - √(x + h)) * (√x + √(x + h))] / (h * √x * √(x + h))
Expanding the numerator:
f'(x) = lim(h->0) [√(x^2) - √(x(x + h)) + √(x(x + h)) - √((x + h)^2)] / (h * √x * √(x + h))
This simplifies to:
f'(x) = lim(h->0) [0] / (h * √x * √(x + h))
Since the numerator is 0, the limit becomes:
f'(x) = lim(h->0) 0 / (h * √x * √(x + h))
Step 5: Simplify the limit:
f'(x) = 0
Therefore, the limit of f'(x) = 1/(√x) as x approaches 0 using the limit definition of derivative is 0.