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

• I
• murshid_islam
In summary: One is ##x+1## and the other is ##\dfrac{x^2-1}{x-1}.## The problem is that ##\dfrac{x^2-1}{x-1}## is undefined at ##x=1## because the denominator is zero. The quotient function is defined to be the difference$$\dfrac{x^2-1}{x-1} - (x+1)$$for ##x \neq 1## and ##0## at ##x=1.## This function is continuous. The definition of ##\lim_{x \to 1}f(x)## is$$\text{For all }\epsilon > #### murshid_islam TL;DR Summary A question about limit $$\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)$$ Apparently, we can divide by $(x-1)$ because $x \neq 1$, but then we plug in $x = 1$ and get 2 as the limit. Is $x = 1$ or $x \neq 1$? What exactly is happening here? murshid_islam said: TL;DR Summary: A question about limit $$\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)$$ Apparently, we can divide by $(x-1)$ because $x \neq 1$, but then we plug in $x = 1$ and get 2 as the limit. Is $x = 1$ or $x \neq 1$? What exactly is happening here? You have a line with a tiny gap at ##x=1##. Approaching this gap from left or right sends you to the missing value ##2##. fresh_42 said: You have a line with a tiny gap at ##x=1##. Approaching this gap from left or right sends you to the missing value ##2##. I get that visually. But I'm questioning the algebra. At first, we divide by (x-1) because $x \neq 1$. But then we plug in x = 1. I'm not being able to make sense of that. murshid_islam said: I get that visually. But I'm questioning the algebra. At first, we divide by (x-1) because $x \neq 1$. But then we plug in x = 1. I'm not being able to make sense of that. You do not plug in ##x=1.## You show that any open neighborhood around ##(1,2)\in \mathbb{R}^2## contains a point of the line, no matter how small this neighborhood is. ##x \neq 1##. The function x+1 happens to be a continuous function f(x) of $$\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$$ But if you plug in x=1, y(1)+1=0+1=1 ##\neq##2 Last edited: murshid_islam said: I get that visually. But I'm questioning the algebra. At first, we divide by (x-1) because $x \neq 1$. But then we plug in x = 1. I'm not being able to make sense of that. It's a bit subtle. You are computing a limit. But, once you have reduced the function in the limit to something continuous and unproblematic at ##x =1## then a quick way to compute the limit is just to plug in ##x =1##. In any case, it should be obvious that$$\lim_{x \to 1} (x +1) = 2$$PS ##\delta = \epsilon## would do the trick if you wanted to prove that from first principles. PeroK said: It's a bit subtle. You are computing a limit. But, once you have reduced the function in the limit to something continuous and unproblematic at ##x =1## then a quick way to compute the limit is just to plug in ##x =1##. My question is - why can we divide the numerator and denominator by ##(x - 1)## by considering ## x \neq 1## and then in the next step consider the opposite, that is, ##x = 1##. That part is not clear to me. Could you explain a bit more? murshid_islam said: My question is - why can we divide the numerator and denominator by ##(x - 1)## by considering ## x \neq 1## and then in the next step consider the opposite, that is, ##x = 1##. That part is not clear to me. Could you explain a bit more? Probably not any more clearly that I did in post #6. What, precisely, do you not understand about what I said in post #6? PeroK said: What, precisely, do you not understand about what I said in post #6? Why can we consider ##x \neq 1## in one step and ##x = 1## in the next step? murshid_islam said: Why can we consider ##x \neq 1## in one step and ##x = 1## in the next step? That's not something I said. I asked: PeroK said: What, precisely, do you not understand about what I said in post #6? Note the word "precisely". PeroK said: That's not something I said. I asked: Note the word "precisely". Then I didn't understand this part: PeroK said: You are computing a limit. But, once you have reduced the function in the limit to something continuous and unproblematic at ##x =1## then a quick way to compute the limit is just to plug in ##x =1##. murshid_islam said: Then I didn't understand this part: Okay. Please (by any means at your disposal) compute:$$\lim_{x \to 1} (x +1)$$##f(x):=\dfrac{(x-1)(x+1)}{(x-1)}## is not defined at ##x=1## so it is forbidden to talk about ##f(1).## ##g(x):=x+1## is defined at ##x=1## so we can build ##g(1)=2.## We want to know whether ##\displaystyle{\lim_{x \to 1}}f(x)## exists. What we also know is ##\displaystyle{\lim_{x \to x_0}}f(x)=g(x_0)## for all ##x_0\in \mathbb{R}.## Hence, for ##x_0=1##$$
\lim_{x \to 1}f(x)=g(1)=2.
$$We plugin ##1## into ##g##, not into ##f.## murshid_islam ... here comes the cavalry! PeroK said: Okay. Please (by any means at your disposal) compute:$$\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. fresh_42 said: 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? PeroK said: ... here comes the cavalry! Do you mean the one from posts #2 and #4 or the cavalry in post #6? murshid_islam said: 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##. murshid_islam said: 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 PeroK said: 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? murshid_islam said: 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. murshid_islam said: 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} = 2murshid_islam and fresh_42 fresh_42 said: murshid_islam said: fresh_42 said: 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. 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).##

murshid_islam said:
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
jbriggs444 said:
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..."

Mark44 said:
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...

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.

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.