Intuition behind calculating limits: why multiply by the conjugate, such as sqrt{t^2+9} + 3?

[jpanknin]

I'm trying to understand the "why" behind this process. We were given the following example:
[math]\lim_{x \to 0} \frac{\sqrt{t^2+9} - 3}{t^2}[/math]
We then proceeded through the calculations as follows (omitting the [imath]\lim_{x \to 0}[/imath] for brevity):
[math]\frac{\sqrt{t^2+9} - 3}{t^2} *\frac{\sqrt{t^2+9} + 3}{\sqrt{t^2+9} + 3}[/math][math]\frac{t^2+9-9}{t^2(\sqrt{t^2+9}+3)}[/math][math]\frac{t^2}{t^2(\sqrt{t^2+9}+3)}[/math][math]\frac{1}{\sqrt{t^2+9}+3}[/math][math]\frac{1}{\sqrt{9}+3} = \frac{1}{6}[/math]
My question is, "what are we doing here?" My confusion is not about this specific problem or about the concept of limits. I'm just using it as an example. There are dozens of others I could have used that all require the same simplification process. My question is about why this process works when finding any limit.

Is the process just about simplifying the original expression? If it is just about simplification, wouldn't taking t to 0 give the correct limit if applied to the original expression too? What does working through this process help us understand about the limit? I know it's giving the right answer, but unsure as to the reasoning behind it.

 

[The Highlander]

Is the process just about simplifying the original expression? If it is just about simplification, wouldn't taking t to 0 give the correct limit if applied to the original expression too? What does working through this process help us understand about the limit? I know it's giving the right answer, but unsure as to the reasoning behind it.

You can't be "taking t to 0" or you would have division by zero! ?

I don't know if that's the sole rationale for adopoting this process; perhaps someone else may be able to advise further?

 

[nasi112]

I'm trying to understand the "why" behind this process. We were given the following example:
[math]\lim_{x \to 0} \frac{\sqrt{t^2+9} - 3}{t^2}[/math]
We then proceeded through the calculations as follows (omitting the [imath]\lim_{x \to 0}[/imath] for brevity):
[math]\frac{\sqrt{t^2+9} - 3}{t^2} *\frac{\sqrt{t^2+9} + 3}{\sqrt{t^2+9} + 3}[/math][math]\frac{t^2+9-9}{t^2(\sqrt{t^2+9}+3)}[/math][math]\frac{t^2}{t^2(\sqrt{t^2+9}+3)}[/math][math]\frac{1}{\sqrt{t^2+9}+3}[/math][math]\frac{1}{\sqrt{9}+3} = \frac{1}{6}[/math]
My question is, "what are we doing here?" My confusion is not about this specific problem or about the concept of limits. I'm just using it as an example. There are dozens of others I could have used that all require the same simplification process. My question is about why this process works when finding any limit.

Is the process just about simplifying the original expression? If it is just about simplification, wouldn't taking t to 0 give the correct limit if applied to the original expression too? What does working through this process help us understand about the limit? I know it's giving the right answer, but unsure as to the reasoning behind it.

This process will fill the gaps of the points of discontinuity to make the function continuous at these points. In other words, the function \(\displaystyle \frac{\sqrt{t^2+9} - 3}{t^2}\) is not continuous at the point \(\displaystyle t = 0\) while the function \(\displaystyle \frac{1}{\sqrt{t^2+9}+3}\) gives the same graph as well as it is continuous at \(\displaystyle t = 0\).

In fact, \(\displaystyle \frac{\sqrt{t^2+9} - 3}{t^2} \neq \frac{1}{\sqrt{t^2+9}+3}\).

And if you want to manipulate the limit like that, you have to write it like this,

\(\displaystyle \lim_{t \to 0} \frac{\sqrt{t^2+9} - 3}{t^2} = \lim_{t \to 0} \frac{1}{\sqrt{t^2+9}+3}, \ \ \ t \neq 0\)

otherwise, it is a wrong manipulation.

Therefore, mentioning \(\displaystyle t \neq 0\), in every step is very important.

How does this process work?
Well, this is the magic of mathematics. It can gives you a duplicate version of yourself. When you look at it, you know it is not you, but when others look at it, they will think it is you. This is how math works.

And since you have no confusion about the concept of limits, it is up to you how to think of the big picture of limits.

 

[blamocur]

I know it's giving the right answer, but unsure as to the reasoning behind it.

There is a bag of popular tricks which you learn as you do more exercises. E.g., when I see [imath]\sqrt{x} -y[/imath] I have a knee-jerk reaction to multiply it by [imath]\sqrt{x}+y[/imath] to get rid of the square root.

 

[The Highlander]

This process will fill the gaps of the points of discontinuity to make the function continuous at these points. In other words, the function \(\displaystyle \frac{\sqrt{t^2+9} - 3}{t^2}\) is not continuous at the point \(\displaystyle t = 0\) while the function \(\displaystyle \frac{1}{\sqrt{t^2+9}+3}\) gives the same graph as well as it is continuous at \(\displaystyle t = 0\).

In fact, \(\displaystyle \frac{\sqrt{t^2+9} - 3}{t^2} \neq \frac{1}{\sqrt{t^2+9}+3}\).

And if you want to manipulate the limit like that, you have to write it like this,

\(\displaystyle \lim_{t \to 0} \frac{\sqrt{t^2+9} - 3}{t^2} = \lim_{t \to 0} \frac{1}{\sqrt{t^2+9}+3}, \ \ \ t \neq 0\)

otherwise, it is a wrong manipulation.

Therefore, mentioning \(\displaystyle t \neq 0\), in every step is very important.

How does this process work?
Well, this is the magic of mathematics. It can gives you a duplicate version of yourself. When you look at it, you know it is not you, but when others look at it, they will think it is you. This is how math works.

And since you have no confusion about the concept of limits, it is up to you how to think of the big picture of limits.

So does that mean that avoiding division by zero is the sole reason for doing this?

 

[Dr.Peterson]

My question is about why this process works when finding any limit.

Is the process just about simplifying the original expression? If it is just about simplification, wouldn't taking t to 0 give the correct limit if applied to the original expression too? What does working through this process help us understand about the limit? I know it's giving the right answer, but unsure as to the reasoning behind it.

So does that mean that avoiding division by zero is the sole reason for doing this?

In a sense. But it's a little more general than that.

You can't just "plug in" t=0, because the original function is undefined there. So you want to make it possible to do so.

You are trying to rewrite the given function to make a new function that is equal to the original except at the point of interest, but is continuous. If you can do this, then the limit is just the value of the new function at that point. (This is essentially what @nasi112 said.)

This can't always be done; many limits, such as sin(x)/x, need other methods. So this process doesn't work when finding any limit. But it is the first thing to consider for many such problems.

 

[Steven G]

I think the OP needs to simplify these three fractions.
(not zero)/0 = ?

0/(not zero) = ?

0/0 = ?

 

[Steven G]

What you are doing is basically cancelling out 0/0 so that you can then substitute 0 for t.

 

[jpanknin]

I think the OP needs to simplify these three fractions.
(not zero)/0 = ?

0/(not zero) = ?

0/0 = ?

@Steven G

(not zero)/0 = undefined

0/(not zero) = zero

0/0 = undefined (could use L'Hospitals, but we're not there yet)

Your second post about cancelling out 0/0 is more along the lines of what I'm asking. I can work out these problems using the limit laws, etc. (and get them right), but don't really understand why this process works and what we're trying to accomplish by manipulating these expressions from the original expression to the expression that will allow us to take the limit. Said another way, when we look at a problem that asks for the limit of an expression, what goal state of the expression are we trying to achieve to allow us to get the limit as t --> 0?

 

[Steven G]

0/0 is said to be an indeterminate form, not undefined. You answered my questions very nicely. Most students get two out three wrong.

If you have the \(\displaystyle lim_{x \rightarrow 5} \dfrac {x-5}{x-5}\) you can reduce (x-5)/(x-5) to 1 since (x-5)/(x-5) = 1 ONLY IF x is not 5 (it is indetermined when x=5). The reason in the limit we get 1 is since x is approaching 5 and will NEVER be 5.
Does that clarify your concerns?

 

[jpanknin]

@Steven G, thank you.

I found this post on stackexchange that asks the question much more clearly than I did and the answer seems to make sense. It's about manipulating the expression into a form that's continuous:

Once we've got it into the form of a continuous function (by which I roughly mean "you could draw it without taking your pencil off the paper", with some technical difficulties that mess that up) defined on a neighbourhood of the point (by which I mean you don't have any undefined-ness happening at the point, or at a sequence of points converging to it), you can make use of the fact that the limit of a continuous function at a point that it is defined in a neighbourhood of is equal to its value at that point: since ?2+2?+4�2+2�+4 is continuous and defined everywhere (unlike ?3−23?−2�3−23�−2, which is not defined at ?=2�=2), we can just substitute in ?=2�=2 to find the limit.

Full post here: https://math.stackexchange.com/ques...ts-results-change-with-algebraic-manipulation

From my reading of this post and a few others, it seems that direct substitution can be used for continuous functions (polynomials, etc.), but for those with discontinuities the expression must be made into a continuous expression.

Does the 1) simplifying and 2) removing discontinuities explanation make sense? Or is that still an incomplete understanding of the goal?

 

[khansaheb]

@Steven G, thank you.

I found this post on stackexchange that asks the question much more clearly than I did and the answer seems to make sense. It's about manipulating the expression into a form that's continuous:



Full post here: https://math.stackexchange.com/ques...ts-results-change-with-algebraic-manipulation

From my reading of this post and a few others, it seems that direct substitution can be used for continuous functions (polynomials, etc.), but for those with discontinuities the expression must be made into a continuous expression.

Does the 1) simplifying and 2) removing discontinuities explanation make sense? Or is that still an incomplete understanding of the goal?

"....it seems that direct substitution can be used for continuous functions (polynomials, etc.), but for those with discontinuities the expression must be made into a continuous expression."

In response #6, you have learnt that lesson.

You can't just "plug in" t=0, because the original function is undefined there. So you want to make it possible to do so.

 

[Steven G]

@Steven G, thank you.

I found this post on stackexchange that asks the question much more clearly than I did and the answer seems to make sense. It's about manipulating the expression into a form that's continuous:



Full post here: https://math.stackexchange.com/ques...ts-results-change-with-algebraic-manipulation

From my reading of this post and a few others, it seems that direct substitution can be used for continuous functions (polynomials, etc.), but for those with discontinuities the expression must be made into a continuous expression.

Does the 1) simplifying and 2) removing discontinuities explanation make sense? Or is that still an incomplete understanding of the goal?

I never consider about making the function continuous, although that is an acceptable way of thinking about it. I always (did I say ALWAYS) direct substitute first. If I get an indetermined result (0/0) what that means to me is that I need to do more. What does more mean? It means to cancel out 0/0. Then I substitute again, if I don't get 0/0 then I have the answer to the limit, if I get 0/0 again, then I try to cancel out 0/0.....

In my opinion, trying to make the function continuous is overkill, but again it works. The reason I feel it is over kill is because to make the function continuous you need to cancel out 0/0. So why not just think about canceling out 0/0.

I now see why I don't like making the function continuous to take a limit. By the very definition of limit at say x=a, we do NOT care about what is happening at x=a. Rather, we are looking at what is happening when x is near a (but NOT a). You method of making the function continuous will always work but goes against the fact that we don't care at happens at x=a.

As a math person, I do not like your method. However, it will never fail you. So keep on thinking your way and you'll never be wrong. I do request that in the back of you mind that you understand that you don't care about the function value at x=a.

 

[jpanknin]

Thank you, @Steven G. That actually makes a lot of sense to me. Though I would disagree about liking MY method given I didn't even know a method existed in the first place .

 

[khansaheb]

It's about manipulating the expression into a form that's continuous:

You remember that there are two types of discontinuity. This a method where - they are removing the "removable discontinuity" (in case you forgot - dust up your old algebra book where functions are discussed). Sometimes it is cleverly referred as "multiplying by 1".

 

[jpanknin]

You remember that there are two types of discontinuity. This a method where - they are removing the "removable discontinuity" (in case you forgot - dust up your old algebra book where functions are discussed). Sometimes it is cleverly referred as "multiplying by 1".

I thought I had it figured out. Mike Schmidt (baseball player for the Phillies in the 70s/80s) once said about baseball, "Any time you think you have the game conquered, the game will turn right around and punch you in the nose.”

I often feel the same way about math.

 

[Steven G]

I thought I had it figured out. Mike Schmidt (baseball player for the Phillies in the 70s/80s) once said about baseball, "Any time you think you have the game conquered, the game will turn right around and punch you in the nose.”

I often feel the same way about math.

I remember Mike Schmidt, but not the statement you quoted. It is however true in math as it is true in baseball!

 

[Steven G]

There is a bag of popular tricks which you learn as you do more exercises. E.g., when I see [imath]\sqrt{x} -y[/imath] I have a knee-jerk reaction to multiply it by [imath]\sqrt{x}+y[/imath] to get rid of the square root.

I have an instantaneous reaction to multiply by [imath]\sqrt{x}+y[/imath]