murshid_islam
TL;DR Summary
$$\lim_{x \rightarrow 1} \frac{x^2 - 1}{x-1}$$
For this, we first divide the numerator and denominator by $(x-1)$ and we get
$$\lim_{x \rightarrow 1} (x+1)$$
Apparently, we can divide by $(x-1)$ because $x \neq 1$, but then we plug in $x = 1$ and get 2 as the limit. Is $x = 1$ or $x \neq 1$? What exactly is happening here?

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TL;DR Summary: A question about limit

$$\lim_{x \rightarrow 1} \frac{x^2 - 1}{x-1}$$
For this, we first divide the numerator and denominator by $(x-1)$ and we get
$$\lim_{x \rightarrow 1} (x+1)$$
Apparently, we can divide by $(x-1)$ because $x \neq 1$, but then we plug in $x = 1$ and get 2 as the limit. Is $x = 1$ or $x \neq 1$? What exactly is happening here?
You have a line with a tiny gap at ##x=1##. Approaching this gap from left or right sends you to the missing value ##2##.

murshid_islam
You have a line with a tiny gap at ##x=1##. Approaching this gap from left or right sends you to the missing value ##2##.
I get that visually. But I'm questioning the algebra. At first, we divide by (x-1) because $x \neq 1$. But then we plug in x = 1. I'm not being able to make sense of that.

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I get that visually. But I'm questioning the algebra. At first, we divide by (x-1) because $x \neq 1$. But then we plug in x = 1. I'm not being able to make sense of that.
You do not plug in ##x=1.## You show that any open neighborhood around ##(1,2)\in \mathbb{R}^2## contains a point of the line, no matter how small this neighborhood is.

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##x \neq 1##. The function x+1 happens to be a continuous function f(x) of
$$\lim_{x\rightarrow a}f(x)=f(a)$$
You made use of this feature to get 2 with a=1, but x ##\neq## 1.

In a case of discontinuous function, say
y(x)=0 for x = 1
y(x)=x otherwise
$$\lim_{x \rightarrow 1}\frac{y^2-1}{y-1}=\lim_{x \rightarrow 1}\ y+1=1+1=2$$
But if you plug in x=1, y(1)+1=0+1=1 ##\neq##2

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I get that visually. But I'm questioning the algebra. At first, we divide by (x-1) because $x \neq 1$. But then we plug in x = 1. I'm not being able to make sense of that.
It's a bit subtle. You are computing a limit. But, once you have reduced the function in the limit to something continuous and unproblematic at ##x =1## then a quick way to compute the limit is just to plug in ##x =1##.

In any case, it should be obvious that
$$\lim_{x \to 1} (x +1) = 2$$PS ##\delta = \epsilon## would do the trick if you wanted to prove that from first principles.

murshid_islam
It's a bit subtle. You are computing a limit. But, once you have reduced the function in the limit to something continuous and unproblematic at ##x =1## then a quick way to compute the limit is just to plug in ##x =1##.
My question is - why can we divide the numerator and denominator by ##(x - 1)## by considering ## x \neq 1## and then in the next step consider the opposite, that is, ##x = 1##. That part is not clear to me. Could you explain a bit more?

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My question is - why can we divide the numerator and denominator by ##(x - 1)## by considering ## x \neq 1## and then in the next step consider the opposite, that is, ##x = 1##. That part is not clear to me. Could you explain a bit more?
Probably not any more clearly that I did in post #6. What, precisely, do you not understand about what I said in post #6?

murshid_islam
What, precisely, do you not understand about what I said in post #6?
Why can we consider ##x \neq 1## in one step and ##x = 1## in the next step?

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Why can we consider ##x \neq 1## in one step and ##x = 1## in the next step?
That's not something I said. I asked:

What, precisely, do you not understand about what I said in post #6?
Note the word "precisely".

murshid_islam
That's not something I said. I asked:

Note the word "precisely".

Then I didn't understand this part:

You are computing a limit. But, once you have reduced the function in the limit to something continuous and unproblematic at ##x =1## then a quick way to compute the limit is just to plug in ##x =1##.

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Then I didn't understand this part:
$$\lim_{x \to 1} (x +1)$$

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##f(x):=\dfrac{(x-1)(x+1)}{(x-1)}## is not defined at ##x=1## so it is forbidden to talk about ##f(1).##
##g(x):=x+1## is defined at ##x=1## so we can build ##g(1)=2.##

We want to know whether ##\displaystyle{\lim_{x \to 1}}f(x)## exists. What we also know is ##\displaystyle{\lim_{x \to x_0}}f(x)=g(x_0)## for all ##x_0\in \mathbb{R}.## Hence, for ##x_0=1##
$$\lim_{x \to 1}f(x)=g(1)=2.$$
We plugin ##1## into ##g##, not into ##f.##

murshid_islam
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... here comes the cavalry!

murshid_islam
$$\lim_{x \to 1} (x +1)$$
What I am having trouble understanding is why we can compute ##\lim_{x \rightarrow 1} (x+1)## to get ##\lim_{x \rightarrow 1} \frac{x^2-1}{x-1}## when ##(x+1)## and ## \frac{x^2-1}{x-1}## are different functions.

murshid_islam
We want to know whether ##\displaystyle{\lim_{x \to 1}}f(x)## exists. What we also know is ##\displaystyle{\lim_{x \to x_0}}f(x)=g(x_0)## for all ##x_0\in \mathbb{R}## .
That is exactly what I am not getting. How do we know that those two are equal?

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... here comes the cavalry!
Do you mean the one from posts #2 and #4 or the cavalry in post #6?

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What I am having trouble understanding is why we can compute ##\lim_{x \rightarrow 1} (x+1)## to get ##\lim_{x \rightarrow 1} \frac{x^2-1}{x-1}## when ##(x+1)## and ## \frac{x^2-1}{x-1}## are different functions.
They are the same function, except at ##x = 1##.

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That is exactly what I am not getting. How do we know that those two are equal?
We prove it. It is what makes a singularity a removable one.

It is so obvious that it is not always done, but basically we need to prove it.

murshid_islam
murshid_islam
They are the same function, except at ##x = 1##.
I'm confused again. Why are they the same if there's an exception and the domains are different?

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I'm confused again. Why are they the same if there's an exception and the domains are different?
If you are taking the limit as ##x \to 1##, then the relevent domain in this case is ##\mathbb R - \{1\}##. On that domain the functions are equal.

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What I am having trouble understanding is why we can compute ##\lim_{x \rightarrow 1} (x+1)## to get ##\lim_{x \rightarrow 1} \frac{x^2-1}{x-1}## when ##(x+1)## and ## \frac{x^2-1}{x-1}## are different functions.
Because we know that ##(x+1) = \frac{x^2-1}{x-1}## for all values of ##x## except ##x=1##. So they are not that different.

And since ##g(x) = (x+1)## is known at ##x=1## then ##\lim_{x \rightarrow 1} (x+1) = g(1) = ((1)+1) = 2##. Or, in other words, the limit close to ##x=1## is equal to the value at ##x=1##. (But it is still the limit we evaluate.)

Personally, when I saw the problem, I used the l'Hôpital's rule:
$$\lim_{x \rightarrow 1} \frac{x^2-1}{x-1} = \lim_{x \rightarrow 1} \frac{\frac{d}{dx}(x^2-1)}{\frac{d}{dx}(x-1)}= \lim_{x \rightarrow 1} \frac{2x}{x} = 2$$

murshid_islam and fresh_42
murshid_islam
We want to know whether ##\displaystyle{\lim_{x \to 1}}f(x)## exists. What we also know is ##\displaystyle{\lim_{x \to x_0}}f(x)=g(x_0)## for all ##x_0\in \mathbb{R}.##

That is exactly what I am not getting. How do we know that those two are equal?

We prove it. It is what makes a singularity a removable one.

It is so obvious that it is not always done, but basically we need to prove it.

How complicated is the proof? I ask because I'm wondering if it will be accessible at my level.

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You prove that ##\lim_{x \to 1} \dfrac{x^2-1}{x-1}=2,## see post #2.

You could e.g. prove that
\begin{align*}
\lim_{x \to 1^+} \dfrac{x^2-1}{x-1}&= \lim_{n \to \infty}\dfrac{\left(1+\dfrac{1}{n}\right)^2-1}{\left(1+\dfrac{1}{n}\right)-1}=\lim_{n \to \infty} \left(1+\dfrac{1}{n}\right)+1=2
\end{align*}
and the same with ##\lim_{x \to 1^+}## and ##-\dfrac{1}{n}.## Then we finally get
$$\lim_{x \to 1^+}\dfrac{x^2-1}{x-1} = 2 = \lim_{x \to 1^-}\dfrac{x^2-1}{x-1}$$
So expanding ##f(x)## by ##(1,2)## results in the continuous completion ##g(x).##

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How complicated is the proof? I ask because I'm wondering if it will be accessible at my level.
If you go back to the definition of limit, you can see the the limit of a function at a point does not ever make use of the value of the function at that point. Let's pull that definition in here...

Definition: ##\lim_{x \to c}f(x) = L##:

There exists a real number L such that for any ##\epsilon > 0## there is a ##\delta > 0## such that if ##0<|x−c|<δ## then ##|f(x)−L|<ϵ##

In this definition, f(c) is never important since ##0 < |x-c|##.

If f(x) defined using the formula: ##\frac{(x+1)(x-1)}{x-1}## and has a range that excludes 1 and if g(x) is defined using the formula ##x+1## and has a range that includes 1 then it is clear that the limits of the two functions as x approaches one will be identical because the value of either function at zero is irrelevant and the two functions are equal everywhere else.

Then, separately, we can argue that since ##g## is continuous, its limit as x approaches one is equal to its value at one, ##g(1)##

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murshid_islam, fresh_42 and PeroK
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If f(x) defined using the formula: ##\frac{(x+1)(x-1)}{x-1}## and has a range that excludes 1 and if g(x) is defined using the formula ##x+1## and has a range that includes 1 then it is clear that the limits of the two functions as x approaches zero will be identical because the value of either function at zero is irrelevant and the two functions are equal everywhere else.
I'm sure you meant "as x approaches 1..."

When $x\rightarrow 1$, the expression approaches $\frac{0}{0}$. This is a classic example of an indeterminate value. L'Hôpitals rule deals with such indeterminates, at least when the functions are well-behaved.