# Single Variable Calculus I: Using L'Hopital's Rule

• Oct 14th 2011, 05:19 PM
DevilDoc
Single Variable Calculus I: Using L'Hopital's Rule
My question is about the type of approach when evaluating the limit as x approaches infinity versus the limit as x approaches a. I understand that L's Rule is used to evaluate indeterminate forms such as 0/0 and infinity/infinity.

For example, lim as x approaches 1 (x^2 - 1)/(x^2 - x) can be factored and evaluated to 2, but what about something like lim as x approaches 1/2 (6x^2 + 5x - 4)/(4x^2 + 16x - 9)?

I get 0/0, which is an indeterminate. I then find the derivative which is (12x+5)/(8x+16) = 11/20. Is this the right thinking or right methodology? The first time I did the problem, I evaluated it down to 3/2, but that was after the second derivative. I'm thinking that's over kill. What do you think?

These problems are for homework. Feel free to post any problems that you think might help me with this tool. By the way, I'll post more questions concerning evaluating an expression as x approaches infinity in the near future. I'm sure I'll have some questions.

Thanks for your help. I appreciate it.
• Oct 14th 2011, 06:49 PM
Prove It
Re: Single Variable Calculus I: Using L'Hopital's Rule
Quote:

Originally Posted by DevilDoc
My question is about the type of approach when evaluating the limit as x approaches infinity versus the limit as x approaches a. I understand that L's Rule is used to evaluate indeterminate forms such as 0/0 and infinity/infinity.

For example, lim as x approaches 1 (x^2 - 1)/(x^2 - x) can be factored and evaluated to 2, but what about something like lim as x approaches 1/2 (6x^2 + 5x - 4)/(4x^2 + 16x - 9)?

I get 0/0, which is an indeterminate. I then find the derivative which is (12x+5)/(8x+16) = 11/20. Is this the right thinking or right methodology? The first time I did the problem, I evaluated it down to 3/2, but that was after the second derivative. I'm thinking that's over kill. What do you think?

These problems are for homework. Feel free to post any problems that you think might help me with this tool. By the way, I'll post more questions concerning evaluating an expression as x approaches infinity in the near future. I'm sure I'll have some questions.

Thanks for your help. I appreciate it.

First of all, while I understand and can follow your thinking, your writing of the mathematics is atrocious.

You say "I then found the derivative", but $\displaystyle \displaystyle \frac{12x + 5}{8x + 16}$ is NOT the derivative of $\displaystyle \displaystyle \frac{6x^2 + 5x - 4}{4x^2 + 16x - 9}$. Rather, you should say "I realised that this is a 0/0 indeterminate form, so in order to use L'Hospital's Rule, and evaluated the derivative OF THE NUMERATOR and the derivative OF THE DENOMINATOR and then evaluated the limit."

Also (12x + 5)/(8x + 16) does NOT equal 11/20. But $\displaystyle \displaystyle \lim_{x \to \frac{1}{2}}\frac{12x + 5}{8x + 16} = \frac{11}{20}$

I'm forgiving, but your lecturer might not be.

Second, you would only ever evaluate the second derivatives of the numerator and denominator (i.e. use L'Hospital's Rule twice) if after using L'Hospital's Rule once, you still get a $\displaystyle \displaystyle \frac{0}{0}$ or $\displaystyle \displaystyle \frac{\infty}{\infty}$ indeterminate form. Since after evaluating the derivatives once, you are able to substitute and get a value for the limit, there is no need, and it is completely wrong to, continue with more uses of L'Hospital's Rule.
• Oct 15th 2011, 08:09 AM
DevilDoc
Re: Single Variable Calculus I: Using L'Hopital's Rule
First and foremost, thanks for the reply. Second, thanks for the clarification of the mathematics. It means a lot to me to not only get the symbolism correct, but also the language that supports it. So your post helped me greatly. I'm sure my professor made the distinction between "...evaluating the derivative of the NUMERATOR and of the derivative of the DENOMINATOR..." He's an interesting character and he does explain the math in such way that it does make sense. But just like anything else, mathematics takes hard work, patients, and lots and lots of practice of doing it right, and doing it right over and over again.

Thanks again for your help concerning this matter. Like I said before, it is much appreciated.

- Doc