L'Hopital's Rule - I'm loosing my hair

In summary, the conversation discusses L'Hopital's Rule and applying it to a specific equation involving a limit as x approaches 0. The participants also mention the use of the chain rule and the importance of understanding the Indeterminate Forms that the rule can be applied to. They also provide an example problem and clarify the solution.
  • #1
anderma8
35
0
L'Hopital's Rule - I'm loosing my hair!

Ok, I have the following:

Lim x->0 sqrt(4-x^2) -2 /x

After I change the equation to remove the radicle, I get:

Lim x->0 ((4-x^2)^1/2 - 2)/x

but when I apply the rule the, I'm loosing it I thought I should get:

1/2 ((4-x^2)^-1/2 times 2x)/1

What am I doing wrong?
 
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  • #2
Don't bother to change the equation, I like radicles :D

Top, use chain rule, bottom, easy. Straight forward now.
 
  • #3
The only thing I see wrong with what you got is that the derivative of 4 - x^2 is -2x. But then again, I'm half asleep.
 
  • #4
Thanks for replying! If I use the chain rule, I get:

1/2 sqrt(4-x^2) -2x ... the two's cancel out but here is what I don't get (and this may be the rust showing on my math skills)... the answer shows as: -x/sqrt(4-x^2)/1 which I sort of understand. Why do I want to put the answer as: 1/2 sqrt(4-x^2) -2x with the radicle in the numerator?
 
  • #5
Well I have that

df/dx = 1/2 ( (4-x^2)^(-1/2)(-2x) )/1
= 1/2 (-2x) / (4-x^2)^(-1/2)
= -x / sqrt(4-x^2)
 
Last edited:
  • #6
Hi Everyone,

thanks for the replies... dimensionless - I think I know what I did wrong. As you have, I didn't raise the exponent to -1/2 thus putting the radicle in the denomonator. Now I'm getting the correct answer, but more importantly - I understand what I did wrong. THANKS for the replies!
 
  • #7
Hey - 1 last question... given that this equates to: -x / sqrt(4-x^2) I don't necessarily see that it goes to 0, unless: since -x is small and sqrt(4-x^2) is smaller, then small/smaller = 0?

Follow me?
 
  • #8
As x -> 0, -x -> 0. As x -> 0, sqrt(4 - x^2) -> 2.

Also, be careful about talking about equivalence here. L'Hospital's Rule just says that, if you have f(x)/g(x) such that f(x) -> 0 and g(x) -> 0 as x -> a, then f(x)/g(x) approaches the same limit as f'(x)/g'(x) as x -> a, if said limit exists.
 
  • #9
Mystic, Thanks for the clarification. That is good to note.
 
  • #10
1 more example to make sure I'm understanding this...If I have:

lim x->1 ln x^2/(x^2+1) this can be rewitten as:

lim x->1 (2 ln x)/(x^2+1) hence using the rule, I get:

2 (1/x)/2x which resolves to:

2/2x^2 and since x->1 this equals 1

Am I getting this right?
 
  • #11
anderma8 said:
1 more example to make sure I'm understanding this...If I have:

lim x->1 ln x^2/(x^2+1) this can be rewitten as:

lim x->1 (2 ln x)/(x^2+1) hence using the rule, I get:

2 (1/x)/2x which resolves to:

2/2x^2 and since x->1 this equals 1

Am I getting this right?
Nope, that's not of any Indeterminate Forms. You should note that L'Hopital's rule can only be applied to the 2 Indeterminate Forms [tex]\frac{0}{0}[/tex] and [tex]\frac{\infty}{\infty}[/tex]. In this problem, we have:
[tex]\lim_{x \rightarrow 1} \ln (x ^ 2) = \ln (1 ^ 2) = 0[/tex]
and:
[tex]\lim_{x \rightarrow 1} (x ^ 2 + 1) = 1 ^ 2 + 1 = 2[/tex], so your limit should be:
[tex]\lim_{x \rightarrow 1} \frac{\ln (x ^ 2)}{x ^ 2 + 1} = \frac{0}{2} = 0[/tex]
You are over-complicating the problem. :rolleyes:
 
  • #12
VietDao, I had a typo: it should have read:

lim x->1 ln x^2/(x^2<b>-</b>1) Sorry for the error. I followed what you had and that makes sense so I think I'm getting it albeit slowly. Using the revised problem, I would get 0/0?
 
  • #13
Yes, if it does read:
[tex]\lim_{x \rightarrow 1} \frac{\ln (x ^ 2)}{x ^ 2 - 1}[/tex], then your solution is totally correct. :)
 
  • #14
:-)

Thanks for the assistance everyone!
 

1. What is L'Hopital's Rule?

L'Hopital's Rule is a mathematical tool used to evaluate limits of indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of two functions, f(x) and g(x), as x approaches a certain value, yields an indeterminate form, then the limit of the quotient of their derivatives, f'(x)/g'(x), will be the same as the original limit.

2. How does L'Hopital's Rule relate to hair loss?

L'Hopital's Rule has no direct relation to hair loss. It is simply a mathematical concept that is often used in calculus courses. Hair loss can be caused by a variety of factors, such as genetics, medical conditions, and lifestyle choices.

3. Is L'Hopital's Rule difficult to understand?

L'Hopital's Rule can be challenging to grasp at first, but with practice, it can become easier to understand. It is important to have a strong understanding of basic calculus concepts, such as derivatives and limits, before attempting to use L'Hopital's Rule.

4. When should I use L'Hopital's Rule?

L'Hopital's Rule should only be used when evaluating limits of indeterminate forms. It is not necessary or applicable in all situations. It is also important to note that L'Hopital's Rule should only be used as a last resort, after trying other methods of evaluating the limit.

5. Are there any limitations to L'Hopital's Rule?

Yes, there are some limitations to L'Hopital's Rule. It can only be used for limits of indeterminate forms, so it is not applicable to all limits. Additionally, it is only applicable in certain situations and may not always yield an accurate result. It is important to understand the concept and its limitations before using it in mathematical calculations.

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