Question about Limit: \lim_{x\rightarrow1} (x+1)

[murshid_islam]

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?

 

[fresh_42]

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.

 

[fresh_42]

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.

 

[anuttarasammyak]

$x \neq 1$. 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

 

[PeroK]

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?

 

[PeroK]

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?

 

[PeroK]

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 $x =1$.

 

[PeroK]

Then I didn't understand this part:

Okay. Please (by any means at your disposal) compute:
$$\lim_{x \to 1} (x +1)$$

 

[fresh_42]

$f(x):=\dfrac{(x-1)(x+1)}{(x-1)}$ 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 $f.$

 

[PeroK]

... here comes the cavalry!

 

[murshid_islam]

Okay. Please (by any means at your disposal) compute:
$$\lim_{x \to 1} (x +1)$$

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?

 

[fresh_42]

... here comes the cavalry!

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

 

[PeroK]

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 $x = 1$.

 

[fresh_42]

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]

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?

 

[PeroK]

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.

 

[jack action]

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]

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.

 

[fresh_42]

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 $g(x).$

 

[jbriggs444]

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, $g(1)$

 

[murshid_islam]

Thanks a bunch @fresh_42 , @jack action and @jbriggs444 . Your explanations helped a lot.

 

[Mark44]

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..."

 

[jbriggs444]

I'm sure you meant "as x approaches 1..."

Yes, I'd gone back and had to edit that post a lot to repair a bunch of that. Just one more edit...

 

[MidgetDwarf]

Your book should have a section or paragraph devoted to removal discontinuity. This will you give you an intuitive approach. Jbriggs444 gave the formal why.

 

[Svein]

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.