Considering this limit :

$\lim\limits_{x \to 1} = \begin{cases} x+1, & \text{x≠1} \\[2ex] \pi , & \text{x=1} \end{cases}$.

from the lift :

$\lim\limits_{x \to 1-} (x+1) = 2 $

from the right:

$\lim\limits_{x \to 1+} (\pi) = \pi $

I'm assuming that this limit is not exist since the left hand side limit does not equal the right hand side limit (From my current knowledge).

The Book that i use telling me not what i expected !:

$\lim\limits_{x \to 1} = \begin{cases} x+1, & \text{x≠1} \\[2ex] \pi , & \text{x=1} \end{cases}$ = $\lim\limits_{x \to 1} (x+1) = 2 $

I think i miss some things about Piecewise limits. anyone explain to me why this limit end up with 2 ?

  • 1
    $\begingroup$ You are not calculating the limit from the right correctly. This limit should also be $2$. $\endgroup$ – platty Dec 1 '18 at 9:05

Recall that, according to the definition, when we take the limit $x\to 1$ we are assuming $x\neq 1$, that is

$$\forall \varepsilon>0 \quad \exists \delta>0 \quad \text{such that}\quad \color{green}{\forall x\neq1}\quad|x-1|<\delta \implies|f(x)-2|<\varepsilon$$

therefore since $x\neq 1$

$$\lim\limits_{x \to 1} f(x)=\lim\limits_{x \to 1} (x+1) = 2$$

In other words, the value for the limit at a point is not affected by the value of the function at that point. The function could be also not defined at that point (e.g. $\sin x/x$ as $x \to 0$).

Your evaluation would be correct for the following function

$$g(x)= \begin{cases} x+1, & \text{$x<1$} \\[2ex] \pi & \text{$x\ge1$} \end{cases}$$

  • $\begingroup$ OK, got the idea $\endgroup$ – Ammar Bamhdi Dec 1 '18 at 9:27
  • $\begingroup$ @AmmarBamhdi Well done! Keep always that in mind. Bye $\endgroup$ – user Dec 1 '18 at 9:28

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