91Ó°ÊÓ

We have to calculate the following limit :

$\begin{array}{l}\underset{x→−1}{lim} xâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots â‹\dots \underset{x→4}{lim} xâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \underset{x→\mathrm{Ï€}}{lim} x\\ \underset{x→0}{lim} x^2â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots â‹\dots \underset{x→5}{lim} x^2\underset{x→-2}{lim} x^2\\ \underset{x→0}{lim} (2x−3)â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots â‹\dots \underset{x→1}{lim} (1−x)â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots â‹\dots \underset{x→3}{lim} (3x+1)\end{array}$

$\underset{x→−1}{lim} x$can be calculated by substituting -1 for c,

$\underset{x→−1}{lim} x=−1$

$\underset{x→4}{lim} x$can be calculated by substituting 4 for c,

$\underset{x→4}{lim} x=4$

$\underset{x→\mathrm{π}}{lim} x$can be calculated by substituting $\mathrm{π}$for c,

$\underset{x→\mathrm{π}}{lim} x=\mathrm{π}$

$\underset{x→0}{lim} x^2$can be calculated by substituting 0 for c,

$\begin{array}{c}\underset{x→0}{lim} x^2=0^2\\ =0\end{array}$

$\underset{x→5}{lim} x^2$ can be calculated by substituting 5 for c,

$\begin{array}{c}\underset{x→5}{lim} x^2=5^2\\ =25\end{array}$

$\underset{x→−2}{lim} x^2$ can be calculated by substituting -2 for c,

$\begin{array}{r}\underset{x→−2}{lim} x^2=(−2{)}^2\\ =4\end{array}$

$\underset{x→0}{lim} (2x−3)$ can be calculated by substituting 0 for c,

$\begin{array}{r}\underset{x→0}{lim} (2x−3)=(2â‹\dots 0−3{)}^2\\ =9\end{array}$

$\underset{x→1}{lim} (1−x)$ can be calculated by substituting 1 for c,

$\begin{array}{c}\underset{x→1}{lim} (1−x)=(1−1)\\ =0\end{array}$

$\underset{x→3}{lim} (3x+1)$ can be calculated by substituting 3 for c,

$\begin{array}{c}\underset{x→3}{lim} (3x+1)=(3â‹\dots 3+1)\\ =9+1\\ =10\end{array}$