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Consider the following function:

$\displaystyle f(x) = \frac{x^2-4}{x-2}$

and now its limit:

$\displaystyle \lim \atop x \to 2$ $\displaystyle \frac{x^2-4}{x-2}$

Upon evaluating, this limit becomes:

$\displaystyle \lim \atop x\to 2$ $\displaystyle x+2 = 4$

The limit therefore exists since it meets each of the following criteria:

The limit approaches a finite value ($4$), the limits on either side of $x=2$ correspond and the limit approaches a particular value ($4$)

With this in mind, does that mean that $\displaystyle \lim_{x\to 0}\frac{1}{x}$ does not exist? Seeing as it does not approach a finite value?


marked as duplicate by Dietrich Burde, Jack, Dando18, Siong Thye Goh, Shailesh Aug 21 '17 at 0:34

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    $\begingroup$ Often limits are not required to be finite. If that criterion were relaxed, what about the limits from the left and the right? $\endgroup$ – sharding4 Aug 20 '17 at 19:52
  • $\begingroup$ Definition, definition, definition. $\endgroup$ – Jack Aug 20 '17 at 20:41
  • $\begingroup$ Roughly your observation is correct. The graph of $y = 1/x$ is not bounded near $x = 0$, so you cannot expect that $\lim_{x\to0} 1/x$ converges in $\mathbb{R}$. But the notion that a limit exists includes not only convergence but also divergence either to $\infty$ or to $-\infty$ (or more formally, convergence with respect to the topology of the extended real line $[-\infty, \infty]$). Of course, $\lim_{x\to 0} 1/x$ does not exist neither as it is unbounded in both ways. $\endgroup$ – Sangchul Lee Aug 20 '17 at 21:17

As you stated in your question

The limit approaches a finite value ($4$), the limits on either side of $x=2$ correspond and the limit approaches a particular value ($4$)

Look at the limit on either side of 0, from the negative side we have $-\infty$ from the other side we have $+\infty$, since the limits from either side are not the same the limit does not exit. However if we had $\lim\limits_{x\to 0^-}\dfrac{1}{x}=-\infty$. Even though this value is infinite, it is still the value of the limit

  • $\begingroup$ Note that the statement $\lim_{x\to a}f(x) = \infty$ has a different definition than $\lim_{x\to a}f(x) = L$, making it, technically, a completely different statement (although the intuition is more or less the same for the two, which is reflected in the notation being practically identical). $\endgroup$ – Arthur Aug 20 '17 at 19:53
  • $\begingroup$ So, what this is saying is that the limit of $\frac{1}{x}$ as $x$ approaches $0$ does not exist when evaluation the limit in a 'double-sided' fashion, but when evaluation the limit in a 'one-sided' fashion, it does exist? $\endgroup$ – joshuaheckroodt Aug 20 '17 at 19:56
  • $\begingroup$ @joshuaheckroodt exactly. For a limit to exist it must approach the same thing from either side. $\endgroup$ – Teh Rod Aug 20 '17 at 20:03
  • $\begingroup$ youtube.com/watch?v=BRRolKTlF6Q $\endgroup$ – user451844 Aug 20 '17 at 20:32

There is a useful proposition which states that the limit of a function(and a sequence for instance) if it exists then it is unique.

Thus uniqueness is a necessary condition for a limit of a function to exist as $x$ aproaches a certain value.

In other words you can think that if a function gives two different limits when its input aproaches a certain value then the limit does not exist.

Thus the limit if it exists will be independent tha path we chose to approach it

(In the real line we have only two paths,from right and left and for instance in the plane we have infinitely many paths when we have a function of two variables.)

Also the existence of the limit does not consern only limits that are real numbers.

In some books if a function has plus infinity for instance as its input aproaches a value then the author can say and says that the limit exists and its infinity.

Take for instance the function you wrote in your post.It does not have a limit at all neither plus or minus infinity


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