Use the Intermediate Value Theorem

scorpa · Sep 27, 2005

Hello everyone,

Unfortunately, my calculus class has just gotten to my least favorite subject in the world.....limits. There are a few questions I am hopelessly stuck on. Any help would be appreciated.

The limit as x approaches 1 of (square root x -x^2)/(1-square root x)

I have no idea how to go about this one. I know that usually when you have a square root you rationalize the equation, but when you have one in the denominator and numerator what do you do?

Use the Intermediate Value Theorem toshow that there is a root of the given equation in the specified interval. Tanx=2x, (0,1.4)

For this one I changed it into the form f(x)=tanx-2x , then I found that f(0)=0 and f(1.4)=2.99 therefore f(0) < 0 < f(2). Is this all I have to do, Is this enough to satisfy the question asking if there is a root between (0,1.4)???

My next questions are ones I was able to get answers for but am unsure of whether I did them correctly.

Find the limit as x->2 of (2x^2+1)/(x^2+6x-4).

For this one I believe you are able to simply substitute 2 in for x and that will give you an answer of 3/4

Find the lim as x ->1 of (x^3-1)/(x^2-1)

For this question I factored out (x-1) and from there I was able to get an answer of (3/2).

My other questions are similar, and if I know I did these right I can be fairly sure the others are correct. Thank you so much for your time and help.

Tzar · Sep 27, 2005

For the first question have you tried L'Hopitals rule?

scorpa · Sep 28, 2005

I'm not aware of that rule, could you elaborate on it? Everytime I try the question I get that the limit is undefined, could it be possible that the limit just does not exist?

VietDao29 · Sep 28, 2005

In #1,
[tex]\lim_{x \rightarrow 1} \frac{\sqrt{x - x ^ 2}}{1 - \sqrt{x}}[/tex]
You can get rid of the square root in the denominator by multiplying both numerator and denominator with [itex]1 + \sqrt{x}[/itex], so the denominator will become 1 - x. Just do it and see what you get.
Or you can try:
[tex]\lim_{x \rightarrow 1} \frac{\sqrt{x - x ^ 2}}{1 - \sqrt{x}} = \lim_{x \rightarrow 1} \frac{\sqrt{x}\sqrt{1 - x}}{1 - \sqrt{x}} = \lim_{x \rightarrow 1} \frac{\sqrt{x}\sqrt{(1 - \sqrt{x})(1 + \sqrt{x})}}{1 - \sqrt{x}}[/tex]
Can you go from here?
Viet Dao,

TD · Sep 28, 2005

scorpa said:

I'm not aware of that rule, could you elaborate on it? Everytime I try the question I get that the limit is undefined, could it be possible that the limit just does not exist?

That's possible but not necessarily so.

L'Hopital's rule says that when you encounter the undeterminate forms 0/0 or [itex]\infty /\infty[/itex], you're allowed to take (separately!) the derivative of nominator and denominator. So if the limit f(x)/g(x) yields such a form, you may replace it by f'(x)/g'(x).

scorpa · Sep 28, 2005

Thanks guys I think I got it from here!!! I really appreciate it!

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Use the Intermediate Value Theorem

scorpa

Tzar

scorpa

VietDao29

TD

scorpa

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