Before I ask, I want to tell you that I am beginner in limits, so you may find some problems in my understanding.

Let's assume a function $f(x) = 15-2x^2$. We want to know how the function behaves at $x=1$. Specifically, we want to know the slope of tangent line at $x=1$.

Simply, we get a good formula for that by doing this: $$m=\frac{f(1)-f(x)}{x-1}.$$ Then we get the equation $$m=\frac{2-2x^2}{x-1}.$$

Now we have to take the limit to find the slope of the tangent line, $$\lim_{x\to 1} \frac {2-2x^2}{x-1}.$$

To solve this we simplify it like this : \begin{align} \lim_{x\to 1} \frac {2-2x^2}{x-1} =& \lim_{x\to 1}\frac {-2(x-1)(x+1)}{(x-1)} \\ =& \lim_{x\to 1}-2(x+1) \\ =& -2(1)-2 \\ =& -4 \end{align}

In algebra class when we had a fraction and we wanted to cancel something we always say $x \ne a$. For instance $\frac {1}{x-1}$. Here $x \ne 1$, because $x-1$ would be zero.

But here in limits I found something unbelievable: here we are dividing by zero and that's forbidden.

$$\lim_{x\to 1}\frac {-2(1-1)(1+1)}{(1-1)}.$$

We are just canceling zero in this fraction. Can anyone explain this?

  • 1
    $\begingroup$ Simply put, we don't. These expressions are not $0$ but they approach $0$. There's a difference. $\endgroup$ – Paras Khosla Apr 12 '19 at 16:56
  • $\begingroup$ The limit is about what happens near $x=1$, not at $x=1$. At $x=1$, the expression you are working with is undefined. However, at any point near $x=1$, the function is perfectly well defined. $\endgroup$ – Xander Henderson Apr 12 '19 at 16:56
  • $\begingroup$ @ParasKhosla , But we plug 1 ! $\endgroup$ – Mohammad Alshareef Apr 12 '19 at 16:57
  • $\begingroup$ Yes because for $x-1$ and $x\to 1$ that's what it approaches. $\endgroup$ – Paras Khosla Apr 12 '19 at 17:00
  • $\begingroup$ Why we don't plug number that approach from $1$.Always $1=1$ not approaching from $1$ $\endgroup$ – Mohammad Alshareef Apr 12 '19 at 17:05

Intuitively, what the limit is doing is finding the behavior of the fraction $$\frac{-2(x-1)(x+1)}{(x-1)}$$ as $x$ approaches $1$. The actual value of the fraction at $1$ is irrelevant, and in this case undefined. Therefore you can 'cancel' out the terms $(x-1)$ because everywhere other than $x = 1$ $$\frac{-2(x-1)(x+1)}{(x-1)} \quad \text{and}\quad -2(x+1)$$ are equal, and so their limit as $x$ approaches $1$ will also be equal.

Another way to think about it is that the function $$f(x) = \frac{-2(x-1)(x+1)}{(x-1)}$$ has a hole in the graph at $x = 1$, whereas the graph of the function $$g(x) = -2(x+1)$$ looks exactly the same except that the hole has been filled in.


You are actually canceling the factor $x-1$ from numerator and denominator. This works as long as $x \ne 1$. Keep in mind that in the limit, $x$ is approaching $1$; never actually equal to $1$.

  • $\begingroup$ Ok , But we at last we plug 1 in the equation ! $\endgroup$ – Mohammad Alshareef Apr 12 '19 at 16:58
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    $\begingroup$ @MohammadAlshareef : That's ONLY because after the cancellation, what remains is a function $f$ that is continuous at $x=1$. It is only by continuity of $f$ that you can then say that $\lim\limits_{x\to1}f(x)=f(1)$. Before you cancel, the original expression is not continuous at $x=1$ (it isn't even defined there!), so it is not allowed to plug in $x=1$. $\endgroup$ – MPW Apr 12 '19 at 17:01
  • $\begingroup$ Cancellation is forbidden , I am with you that it approach from one and not equal one, But we plug one , Specifically , We say that it approach from one then we plug one in the limit ! $\endgroup$ – Mohammad Alshareef Apr 12 '19 at 17:18

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