Evaluate the limit x-->1 (sqrt x - x^2) /(1- sqrt x)

[Nicole Hawley]

I need help with evaluating the limit: lim x--> 1 √x - x²
1- √x

I've tried factoring the numerator to get 1- √x and cancel it out, but can't find a way to factor the numerator to get 1- √x. I've tried rationalizing the denominator, and I'm also tried rationalizing the numerator. Nothing works. I graphed the function, and based on the graph, I know the limit exists. Any suggestions?

 

[tkhunny]

I need help with evaluating the limit: lim x--> 1 √x - x²
1- √x

I've tried factoring the numerator to get 1- √x and cancel it out, but can't find a way to factor the numerator to get 1- √x. I've tried rationalizing the denominator, and I'm also tried rationalizing the numerator. Nothing works. I graphed the function, and based on the graph, I know the limit exists. Any suggestions?

You have written \(\displaystyle \dfrac{\sqrt{x} - x^{2}}{1-\sqrt{x}}\). Is this your intent?

 

[Nicole Hawley]

You have written \(\displaystyle \dfrac{\sqrt{x} - x^{2}}{1-\sqrt{x}}\). Is this your intent?

Yes, that's correct!

 

[mmm4444bot]

Have you learned about L’Hospital’s Rule?

When substitution leads to 0/0 or ∞/∞, then:

 

[Nicole Hawley]

Thank you for that suggestion- yes, I am familiar with L'Hospital's Rule. But this problem is in the textbook well before it introduces L'Hospital's Rule, so there must be a way to find the limit algebraically?

Have you learned about L’Hospital’s Rule?

When substitution leads to 0/0 or ∞/∞, then:

 

[Deleted member 4993]

Thank you for that suggestion- yes, I am familiar with L'Hospital's Rule. But this problem is in the textbook well before it introduces L'Hospital's Rule, so there must be a way to find the limit algebraically?

Yes ... there is a way.

Factorize (√x - x²) - in terms of (1-√x) → then

eliminate (1-√x) from the denominator.

 

[Nicole Hawley]

Yes ... there is a way.

Factorize (√x - x²) - in terms of (1-√x) → then

eliminate (1-√x) from the denominator.

But that's the problem... I can't find a way to factor the numerator in terms of (1-√x)

 

[Deleted member 4993]

But that's the problem... I can't find a way to factor the numerator in terms of (1-√x)

Start with a substitution:

u = √x → u² = x → u⁴ = x²

Now continue and show your work...

 

[Nicole Hawley]

Start with a substitution:

u = √x → u² = x → u⁴ = x²

Now continue and show your work...

Using that substitution, the equation becomes:

(u-u⁴)/(1-u)

Recognizing the numerator is the difference of two squares, I could factor:

[(u^1/2 - u²)(u^1/2 + u²)]/ (1-u)

But I don't see how that helps me, and I can't think of any other way to factor to help me, either.

 

[ksdhart2]

Using that substitution, the equation becomes:

(u-u⁴)/(1-u)

Recognizing the numerator is the difference of two squares, I could factor:

[(u^1/2 - u²)(u^1/2 + u²)]/ (1-u)

But I don't see how that helps me, and I can't think of any other way to factor to help me, either.

Actually, I wouldn't use the difference of squares formula. That will only lead you further down the rabbit hole and have you chasing your tail some more. Instead, I'd begin by factoring out a u from the numerator. \(\displaystyle \dfrac{u(1-u^3)}{1-u}\). Then use the difference of cubes formula. Where does that lead you?

 

[Nicole Hawley]

Actually, I wouldn't use the difference of squares formula. That will only lead you further down the rabbit hole and have you chasing your tail some more. Instead, I'd begin by factoring out a u from the numerator. \(\displaystyle \dfrac{u(1-u^3)}{1-u}\). Then use the difference of cubes formula. Where does that lead you?

Oh wow! I've been staring at this problem for the past couple of days, but I don't think I would have thought to factor it that way no matter how long I looked at it. Thank you so much!!