# L'Hospital's Rule Square Root Problem

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• November 9th 2012, 03:41 PM
Jason76
L'Hospital's Rule Square Root Problem
Find the limit as x approaches 1.

Step 1)

Plug in 1 into x.

$\frac{x - 1}{\sqrt{x^{2} + 3} - 2}$

This come out to being indeterminate or 0/0.

So taking the rule:

Step 2). $\frac{1}{\frac{x}{\sqrt{x^{2} + 3}}}$

Step 3) $\frac{\sqrt{x^{2}+ 3}}{x}$ Final Answer is 2.

Anyhow, I see the pattern of how the answer came about, but I don't understand the reasoning behind it. Well, I understand differintiation part. However. where did the fraction inside of the fraction come from in Step 2? Why did the square root in Step 2 get moved to the numerator in the final answer?
• November 9th 2012, 03:47 PM
Plato
Re: L'Hospital's Rule Square Root Problem
Quote:

Originally Posted by Jason76
$\frac{x - 1}{\sqrt{x^{2} + 3} - 2}$
This come out to being indeterminate.

No it does not.
$\frac{x - 1}{\sqrt{x^{2} + 3} - 2}=\frac{(x - 1)(\sqrt{x^{2} + 3} +2)}{x^2-1}$
• November 9th 2012, 03:48 PM
richard1234
Re: L'Hospital's Rule Square Root Problem
@Jason76, what is the derivative of $\sqrt{x^2 + 3} - 2$ with respect to x?
• November 9th 2012, 03:56 PM
Jason76
Re: L'Hospital's Rule Square Root Problem
Quote:

Originally Posted by Plato
No it does not.
$\frac{x - 1}{\sqrt{x^{2} + 3} - 2}=\frac{(x - 1)(\sqrt{x^{2} + 3} +2)}{x^2-1}$

When factored out it comes to 0 in the bottom. However, when not factored it's indeterminate. But either way, L'Hospital's rule is applied.
• November 9th 2012, 04:09 PM
Plato
Re: L'Hospital's Rule Square Root Problem
Quote:

Originally Posted by Jason76
When factored out it comes to 0 in the bottom. However, when not factored it's indeterminate. But either way, L'Hospital's rule is applied.

No it does not. Please learn some basic algebra.
$\frac{(x - 1)(\sqrt{x^{2} + 3} +2)}{x^2-1}=\frac{\sqrt{x^{2} + 3} +2}{x+1}$
• November 9th 2012, 04:18 PM
Jason76
Re: L'Hospital's Rule Square Root Problem
Quote:

Originally Posted by Plato
No it does not. Please learn some basic algebra.
$\frac{(x - 1)(\sqrt{x^{2} + 3} +2)}{x^2-1}=\frac{\sqrt{x^{2} + 3} +2}{x+1}$

In the book, the limit comes out to 2 (via using L' Hospital's rule). In your reasoning (via factoring), it's 3. Plugging 1 into the original equation (without factoring or using the rule), the answer is 0/0 or indeterminate. Something's wrong here.
• November 9th 2012, 05:27 PM
richard1234
Re: L'Hospital's Rule Square Root Problem
Quote:

Originally Posted by Jason76
In the book, the limit comes out to 2 (via using L' Hospital's rule). In your reasoning (via factoring), it's 3. Plugging 1 into the original equation (without factoring or using the rule), the answer is 0/0 or indeterminate. Something's wrong here.

How'd you get 3? Either way I got an answer of 2.

As to your questions in your original post, they can both be answered by differentation, and basic algebra, which you should already know.
• November 9th 2012, 05:31 PM
Plato
Re: L'Hospital's Rule Square Root Problem
Quote:

Originally Posted by Jason76
In the book, the limit comes out to 2 (via using L' Hospital's rule). In your reasoning (via factoring), it's 3. Plugging 1 into the original equation (without factoring or using the rule), the answer is 0/0 or indeterminate. Something's wrong here.

As I said, Please learn to use basic algebra.
• November 9th 2012, 05:52 PM
Jason76
Re: L'Hospital's Rule Square Root Problem
$\frac{x - 1}{\sqrt{x^{2} + 3} - 2}$

$\frac{x - 1}{\sqrt{x^{2} + 3} - 2} \frac{\sqrt{x^{2} + 3} + 2}{\sqrt{x^{2} + 3} + 2}$

$\frac{(x - 1) (\sqrt{x^{2} + 3)})}{x^{2} + 3 - 4}$

$\frac{(x - 1) (\sqrt{x^{2} + 3)})}{x^{2} - 1}$

$\frac{(x - 1) (\sqrt{x^{2} + 3)} + 2)}{(x - 1) (x + 1)}$

$\frac{\sqrt{x^{2} + 3} + 2)}{(x + 1)}$

Plugging in 1 gets 2 for the limit.
• November 9th 2012, 06:00 PM
richard1234
Re: L'Hospital's Rule Square Root Problem
What Plato's saying is, you probably messed up the arithmetic or something when you obtained an answer of 3 (did you forget the square root sign?). Also, what is wrong with the answer being indeterminate and 2? 0/0 is indeterminate because it could equal anything; that's why we use other techniques (L'Hopital's rule or plain algebra) to determine the limit, which is 2.

Let's return back to the original solution, using L'Hopital's rule. Step 2 in your original solution should be easy if you know how to find derivatives. To get to step 3, you should also know that.
• November 9th 2012, 06:04 PM
Jason76
Re: L'Hospital's Rule Square Root Problem
Quote:

Originally Posted by richard1234
What Plato's saying is, you probably messed up the arithmetic or something when you obtained an answer of 3 (did you forget the square root sign?). Also, what is wrong with the answer being indeterminate and 2? 0/0 is indeterminate because it could equal anything; that's why we use other techniques (L'Hopital's rule or plain algebra) to determine the limit, which is 2.

Let's return back to the original solution, using L'Hopital's rule. Step 2 in your original solution should be easy if you know how to find derivatives. To get to step 3, you should also know that.

• November 9th 2012, 06:08 PM
richard1234
Re: L'Hospital's Rule Square Root Problem
Quote:

Originally Posted by Jason76
$\frac{\sqrt{x^{2} + 3})}{(x + 1)}$

Plugging in 1 gets 2 for the limit.

If you plug in 1 with the expression you provided, you get 1. Not 2.

You're missing a "+2" in the numerator.
• November 9th 2012, 06:14 PM
Jason76
Re: L'Hospital's Rule Square Root Problem
Went back and edited the post, it came out to 3, Plato was right. However, the book says 2. If differentiated using L Hospital's rule, then it comes out to 2. The book is from a major bookstore. I guess it's wrong.

Edited Post:

$\frac{x - 1}{\sqrt{x^{2} + 3} - 2}$

$\frac{x - 1}{\sqrt{x^{2} + 3} - 2} [\frac{\sqrt{x^{2} + 3} + 2}{\sqrt{x^{2} + 3} + 2}]$

$\frac{(x - 1) (\sqrt{x^{2} + 3)})}{x^{2} + 3 - 4}$

$\frac{(x - 1) (\sqrt{x^{2} + 3)})}{x^{2} - 1}$

$\frac{(x - 1) (\sqrt{x^{2} + 3)} + 2}{(x - 1) (x + 1)}$

$\frac{\sqrt{x^{2} + 3} + 2}{(x + 1)}$

Plugging in 1 gets 3 for the limit.
• November 9th 2012, 06:23 PM
Plato
Re: L'Hospital's Rule Square Root Problem
Quote:

Originally Posted by Jason76
Went back and edited the post, it came out to 3, Plato was right. However, the book says 2. If differentiated using L Hospital's rule, then it comes out to 2. The book is from a major bookstore. I guess it's wrong.

Edited Post:

$\frac{x - 1}{\sqrt{x^{2} + 3} - 2}$

$\frac{x - 1}{\sqrt{x^{2} + 3} - 2} \frac{\sqrt{x^{2} + 3} + 2}{\sqrt{x^{2} + 3} + 2}$

$\frac{(x - 1) (\sqrt{x^{2} + 3)})}{x^{2} + 3 - 4}$

$\frac{(x - 1) (\sqrt{x^{2} + 3)})}{x^{2} - 1}$

$\frac{(x - 1) (\sqrt{x^{2} + 3)} + 2)}{(x - 1) (x + 1)}$

$\frac{\sqrt{x^{2} + 3} + 2)}{(x + 1)}$

Plugging in 1 gets 3 for the limit.

As I said learn algebra. Now I say Learn basic arithmetic.
$\frac{\sqrt{1^{2} + 3} +2}{1+1}=2$
• November 9th 2012, 06:34 PM
richard1234
Re: L'Hospital's Rule Square Root Problem
Quote:

Originally Posted by Jason76

$\frac{\sqrt{x^{2} + 3} + 2}{(x + 1)}$

Plugging in 1 gets 3 for the limit.

Nope.
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last