# Anyone can explain to me what's the difference between "limit", "limsup" and "liminf" of a function? It would be helpful to explain with concrete example.

Feb 15, 2017

I'll try to give an example below.

#### Explanation:

Example1 :

$f \left(x\right) = \sin \left(\frac{1}{x}\right)$ as $x \rightarrow 0$

Every deleted $\epsilon$ ball around $0$ has supremum $1$, so

${\lim}_{x \rightarrow 0} \text{sup} f \left(x\right) = 1$

Every deleted $\epsilon$ ball around $0$ has infimum $- 1$, so

${\lim}_{x \rightarrow 0} \text{inf} f \left(x\right) = - 1$

As we know ${\lim}_{x \rightarrow 0} \sin \left(\frac{1}{x}\right)$ does not exist.

Example 2:

$g \left(x\right) = x \sin \left(\frac{1}{x}\right)$ as $x \rightarrow 0$

Every deleted $\epsilon$ ball around $0$ has supremum $\epsilon$, so

${\lim}_{x \rightarrow 0} \text{sup} f \left(x\right) = {\lim}_{\epsilon \rightarrow 0} \epsilon = 0$

Every deleted $\epsilon$ ball around $0$ has infimum $- \epsilon$, so

${\lim}_{x \rightarrow 0} \text{inf} f \left(x\right) = {\lim}_{\epsilon \rightarrow 0} - \epsilon = 0$

We know that ${\lim}_{x \rightarrow 0} x \sin \left(\frac{1}{x}\right) = 0$, for two reasons.

We can use the squeeze theorem on the left and right to get the result.

If we have lim sup = lim inf, then that value is also the limit.