# A question about limits

• B4ssHunter
In summary, when finding a limit for an undefined point on a curve, we can reshape the equation without changing anything to find the limit at that point. However, when defining the function at that point, we are limited to using the shape of the equation that the function is defined by. Changing the way the function is expressed may change the function itself. This is illustrated by the example of f(x) = (x+2)/(x^2 - 4) and h(x) = 1/(x-2), where the latter is not the same as the former.

#### B4ssHunter

when we find a limit for an undefined point on a curve , X^2 - 1 / x - 1 at x = 1 for instance
we reshape the equation without actually changing anything to find the limit at this point .
why can't we do that to define the point on the function ?
i mean clearly if we say F(x) = x^2 - 1 / X - 1
we can reshape it by doing F(x) = (x-1) ( X+1 ) / x-1
which is equal to x +1
it only works for the limit , why doesn't it work for defining the function at this point ?
are we only limited to using the shape of the equation that the function is defined by ?
would changing the shape of the function ( the way we express it ) * even though it changes nothing mathematically * change the function ?

We can't divide by zero. That doesn't bother the limit process, since it looks at the function's behaviour at points in the NEIGHBOURHOOD of the pesky point, not on the pesky point itself. We don't divide by zero in the limit process.

THERE, at the pesky point itself, that expression is simply undefined, and useless as a function value.

arildno said:
We can't divide by zero. That doesn't bother the limit process, since it looks at the function's behaviour at points in the NEIGHBOURHOOD of the pesky point, not on the pesky point itself. We don't divide by zero in the limit process.

THERE, at the pesky point itself, that expression is simply undefined, and useless as a function value.

right , so at the point we don't use the expression X^2-1/x-1 when we are discussing a point , but we rather use the expression 1^2-1/1-1 * where 1 is the point* , and thus it is undefined . great
but wait , why don't we even re-express the function so that F(x) = X+1 , now we don't have to divide by zero and it is defined for x = 1 ! ?

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What I've written is basically the difference between f(x)= (x+1)*((x-1)/(x-1)) and g(x)=x+1
Both are also different from h(x)=(x+1)*((x-1)/(x-1))*((x-3)/(x-3))*((x-e)/(x-e))

1 person
oh , so changing the expression of the function is not allowed ? great
thanks :D

B4ssHunter said:
when we find a limit for an undefined point on a curve , X^2 - 1 / x - 1 at x = 1 for instance
Unless you meant x2 - (1/x) - 1, which I'm sure you didn't, use parentheses around the terms in the numerator and the terms in the denominator. Like this: (x2 - 1)/(x - 1)
B4ssHunter said:
we reshape the equation without actually changing anything to find the limit at this point .
why can't we do that to define the point on the function ?
i mean clearly if we say F(x) = x^2 - 1 / X - 1
we can reshape it by doing F(x) = (x-1) ( X+1 ) / x-1
which is equal to x +1
You haven't really done anything. Your new "reshaped" version still has a discontinuity at (1, 2).
B4ssHunter said:
it only works for the limit , why doesn't it work for defining the function at this point ?
are we only limited to using the shape of the equation that the function is defined by ?
would changing the shape of the function ( the way we express it ) * even though it changes nothing mathematically * change the function ?

The graph of f(x) = (x2 - 1)/(x - 1) has what is known as a removable discontinuity at (1, 2). Most books will show the graph as having a "hole" at (1, 2).

You can extend this function to make it continuous at x = 1 by "plugging the hole."
f(x) = (x2 - 1)/(x - 1), if x ≠ 1
f(x) = 2, if x = 1

What is true is this: if f(x)= g(x) for all x except a (or just in some "punctured neighborhood" of a: all x close to a but NOT equal to a).
$$\frac{x^2- 1}{x- 1}= \frac{(x- 1)(x+ 1)}{x- 1}= x+ 1$$
for all x except x= 1 ("for all x NOT equal to a" because the first is not defined at a but the last is). Therefore, $\lim_{x\to 1}\frac{x^2-1}{x- 1}= \lim_{x\to 1} x+ 1= 2$
.
As for "defining the function at this point", yes, you can do this- $\frac{x^2- 1}{x- 1}$ has a "removable discontinuity" at x= 1. If we were to define $$f(x)= \frac{x^2- 1}{x- 1}$$ for x not equal to 1, f(1)= 2, then we have "removed" the discontinuity. But, of course, this is no longer the same function- it is, in fact, exactly the same as the function x+1.

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HallsofIvy said:
What is true is this: if f(x)= g(x) for all x except a (or just in some "punctured neighborhood" of a: all x close to a but NOT equal to a).
$$\frac{x^2- 1}{x- 1}= \frac{(x- 1)(x+ 1)}{x- 1}= x+ 1$$
for all x except x= 1 ("for all x NOT equal to a" because the first is not defined at a but the last is). Therefore, $\lim_{x\to 1}\frac{x^2-1}{x- 1}= \lim_{x\to 1} x+ 1= 2$
.
As for "defining the function at this point", yes, you can do this- $\frac{x^2- 1}{x- 1}$ has a "removable discontinuity" at x= 1. If we were to define $$f(x)= \frac{x^2- 1}{x- 1}$$ for x not equal to 1, f(1)= 2, then we have "removed" the discontinuity. But, of course, this is no longer the same function- it is, in fact, exactly the same as the function x+1.

okay great , now just to make sure
if i am given a function in a certain shape F(X) = (x+2)/(X^2 - 4 ) , for example
i am not allowed to change that shape if i want to find the definition of a point on that function right ?
so F(x) = (x+2) / ( X^2 -4 ) is not the same as H(x) = 1 / ( X - 2 ) , correct ?

B4ssHunter said:
okay great , now just to make sure
if i am given a function in a certain shape F(X) = (x+2)/(X^2 - 4 ) , for example
i am not allowed to change that shape if i want to find the definition of a point on that function right ?
so F(x) = (x+2) / ( X^2 -4 ) is not the same as H(x) = 1 / ( X - 2 ) , correct ?
They are exactly the same except at one point. The graph of F has a hole at (-2, -1/4). The graph of H is defined at that point.

Both graphs have the line x = 2 as a vertical asymptote.

B4ssHunter said:
okay great , now just to make sure
if i am given a function in a certain shape F(X) = (x+2)/(X^2 - 4 ) , for example
i am not allowed to change that shape if i want to find the definition of a point on that function right ?
"Not allowed"? Who's going to stop you?
You are "allowed" to do whatever you want with function- but they won't necessarily be the same when you are finished!

so F(x) = (x+2) / ( X^2 -4 ) is not the same as H(x) = 1 / ( X - 2 ) , correct ?
Yes, that's true. (I wish you wouldn't mix "x" and "X" like that- they are not the same symbol and cannot be assumed to mean the same thing.) F(x) is defined everywhere except at x= 2 and x= -2 while H(x) is defined everywhere except at x= 2.

It is true that "F(x)= H(x) for all x except x= -2". That detail is, unfortunately, often passed over in Calculus classes because, as I said before, since they are the same for all x except x= 2, they have the same limit at x= -2 and that is what we are primarily concerned with.

HallsofIvy said:
"Not allowed"? Who's going to stop you?
You are "allowed" to do whatever you want with function- but they won't necessarily be the same when you are finished! Yes, that's true. (I wish you wouldn't mix "x" and "X" like that- they are not the same symbol and cannot be assumed to mean the same thing.) F(x) is defined everywhere except at x= 2 and x= -2 while H(x) is defined everywhere except at x= 2.

It is true that "F(x)= H(x) for all x except x= -2". That detail is, unfortunately, often passed over in Calculus classes because, as I said before, since they are the same for all x except x= 2, they have the same limit at x= -2 and that is what we are primarily concerned with.

okay , thanks alot
and by not allowed i just mean a change would occur if i do so !

Another thing to remember is that whenever we talk about a function, it is strictly speaking necessary to specify the actual DOMAIN of x's values we choose to work with, and the number set (for example the reals) in which the RANGE of the function values lies.
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Usually, we implicitly think that the domain of x's is what we call the "maximal domain" within the number set the x's belongs to.
In our two cases, we see that the maximal domain of f(x)=(x^2-1)/(x-1) is all reals except x=1, while the maximal domain of g(x)=x+1 is ALL reals.

However, why should we always care about the MAXIMAL domain?

Why can't we regard these functions as being functions on the domain 5<=x<=17?

Note that on THIS domain choice (rather than the choice of maximal domains), we have full equivalence between f(x) and g(x); they are the SAME function, when we look at them as functions from (5,17) into the reals

## 1. What is a limit in mathematics?

A limit in mathematics is a fundamental concept that describes the behavior of a function as its input approaches a certain value. It represents the value that the function approaches as the input gets closer and closer to the specified value.

## 2. How do you calculate limits?

To calculate a limit, you need to evaluate the function at values that are increasingly closer to the specified input value. This can be done using algebraic manipulation, graphing, or various limit rules, such as the limit laws and L'Hopital's rule.

## 3. What is the difference between a one-sided limit and a two-sided limit?

A one-sided limit, also known as a unilateral limit, only considers the behavior of the function from one side of the specified input value. This means that the input values are restricted to either only approaching the specified value from the left or from the right. On the other hand, a two-sided limit, also known as a bilateral limit, considers the behavior of the function from both sides of the input value.

## 4. Can a function have a limit at a point where it is not defined?

Yes, a function can have a limit at a point where it is not defined. This is because a limit only considers the behavior of the function as the input approaches the specified value, not necessarily the value at that point. However, the limit must exist and be the same from both sides of the input value for it to be considered a valid limit.

## 5. What is the importance of limits in calculus?

Limits are crucial in calculus as they are used to define the derivative and integral, which are essential concepts in calculus. They also help us understand the behavior of functions and solve various problems in mathematics, physics, and other fields.