Find the limit as x approaches 1 of (sqrt(x)-x^2)/(1-sqrt(x))

I know finding the limit is quite simple but i have been stuck on this problem for a while now and just can't seem to figure it out.

I know the answer ends up being 3 but i just cant come to that result. Please help

1) Already two other contributors have explained the L-Hospital's rule; as such let me present the alegraic method.

2) Rationalizing the denominator, by multiplying with (1 + √x),

(√x - x²)/(1 - √x) = (√x - x²)(1 + √x)/(1 - x) = {√x - x² + x - x^(5/2)}/(1-x)

= [(√x(1- x²) + x(1 - x)]/(1-x)

= √x(1 + x) + x

As limit x tends to 1, the above = √1(1 + 1) + 1 = 2 + 1 = 3

Thus the value of the limit = 3

l'Hopital's Rule:
lim f(x)/g(x) = lim f'(x)/g'(x)
f(x) = sqrt(x) - x^2
f'(x) = (1/2)x^-(1/2) - 2x
g(x) = 1 - sqrt(x)
g'(x) = -(1/2)x^-(1/2)
lim x->1 of f'(x)/g'(x) = lim x->1 of -1 - 2(1)/[-(1/2)(1)] = -1 + 4 = 3

Now, l'Hopital's Rule does have constraints on it, but they don't apply to this problem i.e., don't just go using the rule unless you know you're not violating any of its conditions.

You could do this problem by using the definition of limit, but I wouldn't suggest it.

Derivative of numerator = (1/2)x^(-1/2) - 2x lim x-->1 = -3/2
Derivative of denominator = -(1/2)x^(-1/2) lim x-->1 = -1/2
By l'Hopital's rule
limit = -3/2 / -1/2 = 3

This problem is tough to factor. Therefore, use L'hopitals rule:

lim x->1 (√x - x^2)/(1 - √x)

lim x->1 (1/(2√x) - 2x)* (-2√x) = -1 + 4*x^(3/2) = 4 - 1 = 3