Finding the Limit of a Fraction with Square Roots | Calculus Homework Help

[Дьявол]

Homework Statement

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

Homework Equations

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.

 

[tiny-tim]

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

Hi Дьявол!

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

if you're not, use a Taylor expansion.

 

[Дьявол]

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

 

[tiny-tim]

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

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?

 

[Дьявол]

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?

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

 

[tiny-tim]

oops!

But after substituting I get the things messed up

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

 

[Дьявол]

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

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

 

[tiny-tim]

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

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

 

[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.

 

[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 $x\to0$.

 

[tiny-tim]

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 $x\to0$.

ooh, that's much better!

nice one, Avodyne! ​

 

[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.

 

[tiny-tim]

@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.

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

 

[Дьявол]

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

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

 

[tiny-tim]

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

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

work it out!

 

[Mark44]

Homework Statement

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

Homework Equations

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.

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.

 

[tiny-tim]

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.

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!

 

[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?

 

[Mark44]

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!

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).

 

[tiny-tim]

= 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?

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

Avodyne multiplied by something that is 1 in the limit. I mulitplied by something that is 1 (taking domain into account).

ah, yes, but that's what I prefer about his method … it's more compact!

 

[Дьявол]

Haha. Ok, now I understood

You must have worked out for about 20 years with Maths to know all this stuff. It is pretty complicated. Also Mark44's way is great. All of you are great!

Anyway, I still can't figure how somebody find out the formula for cosA-cosB or some of the other sum-to-product identities.

 

[tiny-tim]

Anyway, I still can't figure how somebody find out the formula for cosA-cosB or some of the other sum-to-product identities.

If P = A+B/2 and Q = A-B/2, then A = P+Q/2 and B = P-Q/2 …

so write cosA - cosB in terms of P and Q, and use the usual trig formula (same method works for cosA + cosB, and for sinA ± sinB)

anyway, you should learn all these formulas!