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

[anderma8]

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?

 

[Gib Z]

Don't bother to change the equation, I like radicles :D

Top, use chain rule, bottom, easy. Straight forward now.

 

[Mystic998]

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.

 

[anderma8]

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?

 

[dimensionless]

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)

 

[anderma8]

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!

 

[anderma8]

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?

 

[Mystic998]

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.

 

[anderma8]

Mystic, Thanks for the clarification. That is good to note.

 

[anderma8]

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?

 

[VietDao29]

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 $$\frac{0}{0}$$ and $$\frac{\infty}{\infty}$$. In this problem, we have:
$$\lim_{x \rightarrow 1} \ln (x ^ 2) = \ln (1 ^ 2) = 0$$
and:
$$\lim_{x \rightarrow 1} (x ^ 2 + 1) = 1 ^ 2 + 1 = 2$$, so your limit should be:
$$\lim_{x \rightarrow 1} \frac{\ln (x ^ 2)}{x ^ 2 + 1} = \frac{0}{2} = 0$$
You are over-complicating the problem.

 

[anderma8]

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?

 

[VietDao29]

Yes, if it does read:
$$\lim_{x \rightarrow 1} \frac{\ln (x ^ 2)}{x ^ 2 - 1}$$, then your solution is totally correct. :)

 

[anderma8]

:-)

Thanks for the assistance everyone!