# Why can we divide by zero in limits?

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?

• Simply put, we don't. These expressions are not $0$ but they approach $0$. There's a difference. – Paras Khosla Apr 12 '19 at 16:56
• 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. – Xander Henderson Apr 12 '19 at 16:56
• @ParasKhosla , But we plug 1 ! – Mohammad Alshareef Apr 12 '19 at 16:57
• Yes because for $x-1$ and $x\to 1$ that's what it approaches. – Paras Khosla Apr 12 '19 at 17:00
• Why we don't plug number that approach from $1$.Always $1=1$ not approaching from $1$ – 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 $$1$$.
• @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 $x=1$. – MPW Apr 12 '19 at 17:01