# Why Limit of $0/x$ is $0$, if $x$ approaches $0$? [duplicate]

It might be a silly question, but to me it is not obvious why the following expression holds:

$$\lim\limits_{x\rightarrow 0}\frac{0}{x}=0 ?$$

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## marked as duplicate by Pedro Tamaroff♦, Davide Giraudo, Hans Lundmark, Kaz, HakimJul 15 '14 at 20:38

Because $\frac0x=0$. – user63181 Jul 15 '14 at 14:08
Can you graph $y=\frac{0}{x}$? It's just $y=0$ but with a hole at $x=0$, which is irrelevant for taking limits. – blue Jul 15 '14 at 14:10
apply L'Hopital's rule? – shortstheory Jul 15 '14 at 14:55
Keep in mind that a limit only describes the behavior of a function as the variable approaches a particular value; it does not have to equal the value of the function at that value of x (if the function is even defined there). – RecklessReckoner Jul 15 '14 at 17:21
It's because for any values of $x$ other than $x = 0$, $\frac{0}{x}$ is zero. Thus no matter how close x gets to zero, the value of the expression is zero, as long as x doesn't reach zero. From this observation it pops out that the limit is zero. – Kaz Jul 15 '14 at 20:27

A limit $L$ of a function $f(x)$ evaluated at a point $x = a$ is not necessarily the value $f(a)$ itself. It is a value for which $f(x)$ approaches $L$ "as close as we like" if we make $x$ "sufficiently close" but not equal to $a$. The subtlety is in how we mathematically formalize the language in quotation marks, which is how we arrived at the Cauchy definition of limit:

We say that $\displaystyle \lim_{x \to a} f(x) = L$ if, for any $\epsilon > 0$, there exists a $\delta > 0$ such that $|f(x) - L| < \epsilon$ whenever $0 < |x - a| < \delta$.

Of course, we do not need to appeal to such a definition in this case because as others have pointed out, $f(x) = 0/x = 0$ whenever $x \ne 0$; hence $$\lim_{x \to 0} \frac{0}{x} = \lim_{x \to 0} 0 = 0$$ directly, because again, the limit is evaluated by considering the behavior of $f(x)$ in a neighborhood of $a = 0$, not the value of $f(0)$.

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$$\frac0{1}=0$$ $$\frac0{0.1}=0$$ $$\frac0{0.01}=0$$ $$\frac0{0.001}=0$$ $$\frac0{0.0001}=0$$ $$\frac0{0.00001}=0$$ $$\frac0{0.000001}=0$$ $$...$$

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Nice visual explanation – Jonathan Landrum Jul 15 '14 at 20:45

It's simple:

The limit is the value that the function approaches at that point, simply put, it depends on the neighboring values the function takes.

Take a graph of the function $f(x)=\frac{0}{x}$:

You see that from any possible angle, the only value the function approaches when $x\rightarrow0$ (or wherever in the known universe) is $0$.

A different scenario would appear with, for example, $f(x)=\frac{sin(x)}{x}$. Here, you can see in the plot that the line approaches $1$ as it gets close to $x=0$, that's why this limit is equal to 1.

And in both cases, $f(0) = \frac{0}{0}$. The function is, in both cases, undefined at that value of $x$, but the limit tells you toward which value it approaches.

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This is true simply because as you take $x \to 0$ (for $x\ne 0$), we have $0/x=0$. (Think about the convergence of the sequence $0,0,0,0,\ldots$.) When you take the limit, you don't care about what happens when $x=0$; you only care about what happens around it.

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You've received great answers thus far, so there's no need to repeat what's already been stated, and stated well.

I'll simply add this, for a different take, given that you've already encountered integrals and double integrals, I take it you've encountered L'Hospital for limits:

Even if you to try to evaluate at the limit at $x = 0,$ we get the indeterminate form $\dfrac 00$. By L'Hospital's rule, then, $$\lim_{x\to 0} \frac 0x = \lim_{x \to 0}\frac{(0)'}{(x)'} = \lim_{x\to 0} \frac 01 = 0.$$

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If somebody doesn't understand such a basic fact about limits, it's probably not the right time to introduce L'Hôpital's rule – Max Jul 15 '14 at 15:08
@Max I looked at other questions asked by the OP, and s/he has certainly been exposed to limits, and integrals, and limits of integrals. (And I'm sure, therefore, as been exposed to L'Hospital.) I suspect the OP had what is sometimes called a momentary "brain fart". In any case, I'm afraid your conclusion is in error. – amWhy Jul 15 '14 at 15:11