How Do You Solve Limits Involving Square Roots as h Approaches Zero?

[KevinFan]

Homework Statement

limit h goes to 0 sqrt(73-2(x+h))-sqrt(73-2x)/h

Homework Equations

The Attempt at a Solution

I am not too sure how to cancel out the "h" in the denominator.
just need some hints for me to start

 

[Ray Vickson]

Homework Statement

https://www.physicsforums.com/file:///C:\Users\Kevin\AppData\Roaming\Tencent\Users\1456235477\QQ\WinTemp\RichOle\VV[{{}L1]ZMRHDNWWCHP9G1.png
limit h goes to 0 sqrt(73-2(x+h))-sqrt(73-2x)/h

Homework Equations

The Attempt at a Solution

I am not too sure how to cancel out the "h" in the denominator.
just need some hints for me to start

You wrote
$$\sqrt{73-2(x+h)} -\frac{\sqrt{73-x}}{h}$$
I hope that is not what you meant; I hope you meant
$$\frac{\sqrt{73-2(x+h)}-\sqrt{73-2x}}{h}$$
If so, you MUST use parentheses, because when you write A-B/C you are writing $A - \frac{B}{C}$; if you want $\frac{A-B}{C}$ you need to write it in plain text as (A-B)/C. See the difference?

 

[KevinFan]

You wrote
$$\sqrt{73-2(x+h)} -\frac{\sqrt{73-x}}{h}$$
I hope that is not what you meant; I hope you meant
$$\frac{\sqrt{73-2(x+h)}-\sqrt{73-2x}}{h}$$
If so, you MUST use parentheses, because when you write A-B/C you are writing $A - \frac{B}{C}$; if you want $\frac{A-B}{C}$ you need to write it in plain text as (A-B)/C. See the difference?

i mean (sqrt(73-2(x+h))-sqrt(73-2x))/h, sorry for any confusion

 

[Ray Vickson]

i mean (sqrt(73-2(x+h))-sqrt(73-2x))/h, sorry for any confusion

What results are you allowed to use? What types of tools are you already equipped with?

 

[KevinFan]

What results are you allowed to use? What types of tools are you already equipped with?

I am only allowed to use exact value as answer and I am equipped with a Casio fx-991 calculator.

 

[Ray Vickson]

I am only allowed to use exact value as answer and I am equipped with a Casio fx-991 calculator.

No, that is not what I mean. I mean what "mental" tools are you allowed to use? How much mathematics do you know already? Have you taken calculus? Have you just started to study calculus? Have you seen similar problems already in your textbook or course notes?

I don't care what type of calculator you are permitted to use; you should PUT AWAY the calculator and work without it! That is the only way to learn this type of material. Save the use of your calculator for doing numerical computations; do algebra using a paper and pencil.

 

[KevinFan]

No, that is not what I mean. I mean what "mental" tools are you allowed to use? How much mathematics do you know already? Have you taken calculus? Have you just started to study calculus? Have you seen similar problems already in your textbook or course notes?

I don't care what type of calculator you are permitted to use; you should PUT AWAY the calculator and work without it! That is the only way to learn this type of material. Save the use of your calculator for doing numerical computations; do algebra using a paper and pencil.

I am taking calculus right now and I am working ahead. We have not covered the material yet but I learned the basic limit laws through my textbook and some videos online.

 

[epenguin]

Well as in a limit problem yesterday, most of your problem here is not a question of limits but rather about algebrical manipulation.
Now you can see that there are some of the same things under the square root signs that you would like to subtract. But roughly speaking you can never just subtract one square root from another. As beginning students often do but you seem to realize you can't. Not because anyone says so but because it is just not logically valid.
The only way to get rid a of square root is to square it! But you can't go around squaring things without a warrant,.That is, your new more convenient expression has to be equal to the old inconvenient one.
Very likely previous algebra you have done dealt with dealing with expressions with square roots.
You might need to look up an old chapter in a previous book.
In other words, you are on a chapter about limits but that doesn't mean that the calculations are all independent of other things you learned before.
I hope this does not seem too cryptic.

Er

After I wrote that I looked at it again. It was not so easy.
What I was saying is that you usually try to deal with these square root problems this way: you say that this expression is the same thing as the square root of the expression squared. With the expression squared you often are able to do something. In this case however it looks like this just gives you a problem like what you started with inside your new expression. If you have tried it you will know what I mean. There might be a way to use this fact of ending up with something like what you started with, but if there is I think it's probably asking too much of a student.

So I think the best way is to use something else you may or may not have studied, That is try an approximate expression for the square root of a thing plus a much smaller thing. That is probably what your textbook will use. The standard form in fact is a square root of one + something very much less than one. (1 + 2a)^½ where |a| <<1 . You have to factor your square root to make it like this last.

(That's the calculation, but about the limit question you'll still need some justification for the approximation being made.)

 

[KevinFan]

Well as in a limit problem yesterday, most of your problem here is not a question of limits but rather about algebrical manipulation.
Now you can see that there are some of the same things under the square root signs that you would like to subtract. But roughly speaking you can never just subtract one square root from another. As beginning students often do but you seem to realize you can't. Not because anyone says so but because it is just not logically valid.
The only way to get rid a of square root is to square it! But you can't go around squaring things without a warrant,.That is, your new more convenient expression has to be equal to the old inconvenient one.
Very likely previous algebra you have done dealt with dealing with expressions with square roots.
You might need to look up an old chapter in a previous book.
In other words, you are on a chapter about limits but that doesn't mean that the calculations are all independent of other things you learned before.
I hope this does not seem too cryptic.

Er

After I wrote that I looked at it again. It was not so easy.
What I was saying is that you usually try to deal with these square root problems this way: you say that this expression is the same thing as the square root of the expression squared. With the expression squared you often are able to do something. In this case however it looks like this just gives you a problem like what you started with inside your new expression. If you have tried it you will know what I mean. There might be a way to use this fact of ending up with something like what you started with, but if there is I think it's probably asking too much of a student.

So I think the best way is to use something else you may or may not have studied, That is try an approximate expression for the square root of a thing plus a much smaller thing. That is probably what your textbook will use. The standard form in fact is a square root of one + something very much less than one. (1 + a)^½ where |a| <<1 . You have to factor your square root to make it like this last.

(That's the calculation, but about the limit question you'll still need some justification for the approximation being made.)

Hi, thank you for your reply! I tried to square the whole thing to get rid of the square root, and I simplified the expression and I got lim h goes 0 (2/h). I noticed I still cannot divide out the "h" in the denominator. However, I also realized that the "h" is not equal to 0. It is just approaching 0. I think in this case a constant divide a number approaching zero is infinity. Is my understanding correct? (I am afraid I don't quite understand the "approximate expression")

 

[Mark44]

Hi, thank you for your reply! I tried to square the whole thing to get rid of the square root, and I simplified the expression and I got lim h goes 0 (2/h).

No, this is not right. It's not valid to "square the whole thing."

The simplest approach to simplifying this:
$$\frac{\sqrt{73 - 2(x + h)} - \sqrt{73 - 2x}}{h}$$
is to multiply the fraction by 1, in the form of the conjugate of the numerator over itself. Multiplication by 1 is always valid, whereas squaring an expression gives you a new expression that is almost always different from the expression you squared.

Here's the basic idea:
$$\frac{\sqrt{a} - \sqrt{b}}{c} = \frac{\sqrt{a} - \sqrt{b}}{c} \cdot \frac{\sqrt{a} +\sqrt{b}}{\sqrt{a} +\sqrt{b}} = \frac{a - b}{c(\sqrt{a} +\sqrt{b})}$$
Although the last fraction looks just as complicated as the first, for your problem, it makes things easier when you take the limit.

I noticed I still cannot divide out the "h" in the denominator. However, I also realized that the "h" is not equal to 0. It is just approaching 0. I think in this case a constant divide a number approaching zero is infinity. Is my understanding correct? (I am afraid I don't quite understand the "approximate expression")

 

[Ray Vickson]

No, this is not right. It's not valid to "square the whole thing."

The simplest approach to simplifying this:
$$\frac{\sqrt{73 - 2(x + h)} - \sqrt{73 - 2x}}{h}$$
is to multiply the fraction by 1, in the form of the conjugate of the numerator over itself. Multiplication by 1 is always valid, whereas squaring an expression gives you a new expression that is almost always different from the expression you squared.

Here's the basic idea:
$$\frac{\sqrt{a} - \sqrt{b}}{c} = \frac{\sqrt{a} - \sqrt{b}}{c} \cdot \frac{\sqrt{a} +\sqrt{b}}{\sqrt{a} +\sqrt{b}} = \frac{a - b}{c(\sqrt{a} +\sqrt{b})}$$
Although the last fraction looks just as complicated as the first, for your problem, it makes things easier when you take the limit.

Just so the OP is not confused even more: that last expression has a typo in it. It should be
$$ \frac{1}{c} \frac{a-b}{\sqrt{a}+\sqrt{b}} $$
Edit (Mark44) -- I fixed it in my post and in the text copied above.

 

[KevinFan]

No, this is not right. It's not valid to "square the whole thing."

The simplest approach to simplifying this:
$$\frac{\sqrt{73 - 2(x + h)} - \sqrt{73 - 2x}}{h}$$
is to multiply the fraction by 1, in the form of the conjugate of the numerator over itself. Multiplication by 1 is always valid, whereas squaring an expression gives you a new expression that is almost always different from the expression you squared.

Here's the basic idea:
$$\frac{\sqrt{a} - \sqrt{b}}{c} = \frac{\sqrt{a} - \sqrt{b}}{c} \cdot \frac{\sqrt{a} +\sqrt{b}}{\sqrt{a} +\sqrt{b}} = \frac{a - b}{c(\sqrt{a} +\sqrt{b})}$$
Although the last fraction looks just as complicated as the first, for your problem, it makes things easier when you take the limit.

Many thanks for your excellent explanation!

 

[epenguin]

½

Hi, thank you for your reply! I tried to square the whole thing to get rid of the square root, and I simplified the expression and I got lim h goes 0 (2/h). I noticed I still cannot divide out the "h" in the denominator. However, I also realized that the "h" is not equal to 0. It is just approaching 0. I think in this case a constant divide a number approaching zero is infinity. Is my understanding correct? (I am afraid I don't quite understand the "approximate expression")

I take back my statement that the only way to get rid of square roots is by squaring. It often is a way when transforming equations, and then you may find you have to do it more than once . But offhand that will give you a mess here, and there is this other way which others have illustrated.

About the 'approximate expression' I mean

(1 + a)² = (1 + 2a + a²)

exactly. Now if a << 1 then a² becomes very very small in comparison with the whole
and (1 + a)² becomes close to (1 + 2a) .

Or, taking square roots, √(1 + 2a) is close to (1 + a). (This is what is used in the calculation of square roots).

Talking about very smalll things ('as small as we please') that in the limit are negligible compared with other things in your expression is part of the standard 'limit' discourse.

The idea is easy to extend to other fractional indices than ½ .