# Lim{x->0} function

1. Nov 19, 2008

### Дьявол

1. The problem statement, all variables and given/known data

Find $$\lim_{x \rightarrow 0}=\frac{\sqrt{x+a^2}-a}{\sqrt{x+b^2}-b}$$

2. Relevant equations

3. The attempt at a solution

$$\lim_{x \rightarrow 0}=\frac{\sqrt{x+a^2}-a}{\sqrt{x+b^2}-b}=\lim_{x \rightarrow 0}=\frac{\sqrt{x+a^2}-a}{\sqrt{x+b^2}-b}*\frac{\sqrt{x+b^2}+b}{\sqrt{x+b^2}+b}=\lim_{x \rightarrow 0}=\frac{(\sqrt{x+a^2}-a)(\sqrt{x+b^2}+b)}{x}$$
On the next step I divided by x, but again nothing.
I tried several different methods by substituting $y=\sqrt{x+a^2}$ and $z=\sqrt{x+b^2}$ but useless. Please help! Thanks in advance.

2. Nov 19, 2008

### tiny-tim

Hi Дьявол!

If you're allowed to, use l'Hôpital's rule …

if you're not, use a Taylor expansion.

3. Nov 19, 2008

### Дьявол

@tiny-tim I have never used both of them. Aren't there any elementary transformations?

4. Nov 19, 2008

### tiny-tim

ok … here's a trick …

whenever I see a √ sign, I think of using a trig substitution, so …

define θ and φ by x = asinθ = bsinφ, and let θ and φ tend to zero, and use standard trigonometric identities …

what do you get?

5. Nov 19, 2008

### Дьявол

Maybe, that is nice trick. If θ and φ tend to 0 then asin0 and bsin0 tend to 0 so that x tend to 0.

But after substituting I get the things messed up

Should I try like this:

$$\frac{\sqrt{asin(\theta)+a^2}-a}{\frac{\sqrt{asin(\theta)+\frac{a^2sin^2\theta}{sin^2(\varphi)}}}{-\frac{asin(\theta)}{sin(\varphi)}}$$
or maybe:
$$\frac{\sqrt{asin(\theta)+a^2}-a}{\sqrt{bsin(\varphi)+b^2}-b}$$

Last edited: Nov 19, 2008
6. Nov 19, 2008

### tiny-tim

oops!

Sorry … I couldn't read my own handwriting

I should have said x = a2tan2θ = b2tan2φ …

or even better x = a2cosθ = b2cosφ, with θ and φ -> π/2

7. Nov 19, 2008

### Дьявол

Re: oops!

Maybe it is better if we use x = -a2cosθ = -b2cosφ (it is the same thing), and I get:
$$\frac{asin^2(\theta)-a}{bsin^2(\varphi)-b}$$
for θ and φ -> π/2
but I can't do anything out of here, or I can (circular movement)

If I do like this:
-a(1-sin2θ/-b(1-sin2φ)=acos2θ/bcos2φ=0/0

8. Nov 19, 2008

### tiny-tim

No, you get (a/b)√(1 - cosθ)/√(1 - cosφ)

9. Nov 19, 2008

### Avodyne

Well, this is all getting very complicated. It's better to learn a more universal method.

The point is, what can we use to approximate $\sqrt{x+a^2}$ when $x$ is small?

In general, answering this sort of question involves using a Taylor expansion,
$$f(x)=f(0)+f'(0)x+{\textstyle{1\over2}}f''(0)x^2+\ldots$$
Typically only the first two terms are needed for a limit problem.

In the case of interest, $f(x)=(x+a^2)^{1/2}$, and so $f'(x)=\frac12 (x+a^2)^{-1/2}$. Thus we have $f(0)=a$ and $f'(0)=1/(2a)$, and so $\sqrt{x+a^2}\simeq a + x/(2a)$ when $x$ is small.

10. Nov 19, 2008

### Avodyne

Here's an alternative method for this particular problem that doesn't use Taylor expansions. Multiply the original expression by
$$\Biggl(\frac{\sqrt{x+a^2}+a}{2a}\Biggr)\Biggl(\frac{2b}{\sqrt{x+b^2}+b}\Biggr)$$
which has a limit of 1 as $x\to0$.

Last edited: Nov 19, 2008
11. Nov 19, 2008

### tiny-tim

ooh, that's much better!

nice one, Avodyne!

12. Nov 19, 2008

### Дьявол

@Avodyne
That's great. I didn't think that I could use something like that, but that's obviously lim1=1; so again I can use it in my particular problem. Thanks a lot.

@tiny-tim
Thanks for the replies. But let's solve it also on your way of thinking.
If x = a2cosθ = b2cosφ, with θ and φ -> π/2
then in the original equation:
√(a2cosθ+a2)-a/ √(b2cosφ+b2)-b=
=a√(1+cosθ)-a/ b√(1+cosφ)-b =a(√(1+cosθ)-1)/b(√(1+cosφ)-1)
What could I do know? Again stuck.

Using Avodyne's method I got b/a which is actual correct, but also I found your way very good and I would like to solve it. Thanks again.

13. Nov 19, 2008

### tiny-tim

Well. it's not pretty

a(√(1+cosθ)-1)/b(√(1+cosφ)-1)

= a((√2)cosθ/2 - 1)/b((√2)cosφ/2 - 1)

= a(cosθ/2 - 1/√2)/b((cosφ/2 - 1/√2)

= a(cosθ/2 - cos45º)/b((cosφ/2 - cos45º)

= a sin(θ/4 + 22.5º) sin(θ/4 - 22.5º) / b sin(φ/4 + 22.5º) sin(φ/4 - 22.5º)

~ a (θ/4 + 22.5º) (θ/4 - 22.5º) / b (φ/4 + 22.5º) (φ/4 - 22.5º)

= a (θ + 90º) (θ - 90º) / b (φ + 90º) (φ - 90º);

using (θ + 90º)/(φ + 90º) ~ 180º/180º = 1

and b2/a2 = cosθ/cosφ = sin(θ - 90º)/sin(φ - 90º) ~ (θ - 90º)/(φ - 90º),​

~ (a/b)(b2/a2) = b/a

14. Nov 19, 2008

### Дьявол

How do you got equal (cosθ/2 - cos45º) = sin(θ/4 + 22.5º) sin(θ/4 - 22.5º) ?

15. Nov 19, 2008

### tiny-tim

Standard trig formula: cosA - cosB = 2 sin(A+B)/2 sin(A-B)/2 …

work it out!

16. Nov 19, 2008

### Staff: Mentor

There's another approach that I didn't see in this thread, that's much simpler than most of the approaches that were suggested:
Multiply the original expression by 1, in the form of
$$\frac{\sqrt{x + a^2} + a}{\sqrt{x + b^2} + b} * \frac{\sqrt{x + b^2} + b}{\sqrt{x + a^2} + a}$$
This makes the limit very easy to evaluate, which when simplified, is b/a.

17. Nov 19, 2008

### tiny-tim

Hi Mark!

Yes, that's nice …

but Avodyne beat you to it in post #10,

and his version was neater 'cos it had less square-roots!

18. Nov 19, 2008

### Дьявол

= a sin(θ/4 + 22.5º) sin(θ/4 - 22.5º) / b sin(φ/4 + 22.5º) sin(φ/4 - 22.5º)

~ a (θ/4 + 22.5º) (θ/4 - 22.5º) / b (φ/4 + 22.5º) (φ/4 - 22.5º)

Sorry for asking, but where the sine gone here?

19. Nov 19, 2008

### Staff: Mentor

I noticed Avodyne's approach and thought that mine was similar to his (hers?), but still different. Avodyne multiplied by something that is 1 in the limit. I mulitplied by something that is 1 (taking domain into account).

20. Nov 19, 2008

### tiny-tim

Magic!

hmm … on reflection , I can only magic away the minus brackets … I have to treat the plus brackets separately …

sin(θ/4 + 22.5º) / sin(φ/4 + 22.5º) ~ sin45º / sin45º = 1,

and

sin(θ/4 - 22.5º) / sin(φ/4 - 22.5º) ~ (θ/4 - 22.5º) / (φ/4 - 22.5º)

because as θ -> 90º, (θ/4 - 22.5º) -> 0, and sinx ~ x as x -> 0
ah, yes, but that's what I prefer about his method … it's more compact!