How Do You Solve This Limit Problem Involving Square Roots and Squares?

[thegame1234]

Immediate help needed (limits)!

hi everyone...I need an answer for this limit question by friday(feb/04/2005)...ok the question is to find the limit as "x" approaches 1 for the equation...sqare root "x" subtract "x" sqared divide by 1 subtract sqare root "x"...sorry guys but i don't know how to use the codes yet...anyways...I used the calculator to find out that the limit is three but I am stuck in doing the algebra. I tried to rationalize both the denominator and the nemoritor and basically came up with (x-4)(1+sqaure root "x")/(1-x)(square root "x" +x^2)...I also multiplied them but couldn't cancel out anything...any help would be greatly appreaciated.

 

[Curious3141]

Substitute u = \sqrt{x} and express everything in terms of u.

For the resulting expression, factorise the numerator completely. Cancel the common factor(s) between numerator and denominator and you get an expression in u that has an obvious limit as u tends to 1.

 

[wais]

Hi there,

Thank you for reaching out for help with your limit question. It looks like you have made some good attempts at solving it already. Let me walk you through the steps to find the limit as x approaches 1 for the given equation.

First, we need to simplify the equation as much as possible. Let's start by factoring the numerator and denominator separately.

Numerator: √x - x^2 = x√x - x^2

Denominator: 1 - √x = 1 - √x

Next, we can factor out an x from both the numerator and denominator.

x(√x - x) / x(1 - √x)

Now, we can cancel out the x's in the numerator and denominator, leaving us with:

√x - x / 1 - √x

Next, we can multiply the numerator and denominator by the conjugate of the denominator, which is 1 + √x.

(√x - x)(1 + √x) / (1 - √x)(1 + √x)

Expanding this out, we get:

√x + x - x√x - x^2 / 1 - √x + √x - x

Simplifying further, we get:

√x - x^2 / 1 - x

Now, we can plug in x = 1 to find the limit:

√1 - 1^2 / 1 - 1 = 0/0

Since we get an indeterminate form, we can use L'Hopital's rule to find the limit. This rule states that if we have an indeterminate form of 0/0 or ∞/∞, we can take the derivative of the numerator and denominator separately and then evaluate the limit again.

Taking the derivatives, we get:

(1/2)(x)^(-1/2) - 2x / -1

= 1/(2√x) - 2x

Now, plugging in x = 1, we get:

1/[2(√1)] - 2(1) = 1/2 - 2 = -3/2

Therefore, the limit as x approaches 1 for the given equation is -3/2.

I hope this helps! Let me know if you have any further questions