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Let $[x]$ denote the greatest integer less than or equal to $x$ and $f(x)=2 x-[2 x]$. If $\lim _{x \rightarrow 2^{-}} f(x)=l_1$ and $\lim _{x \rightarrow 2^{+}} f(x)=l_2$ then $l_1+l_2=$
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Verified Answer
The correct answer is:
1
Given $\mathrm{f}(\mathrm{x})=2 \mathrm{x}-[2 \mathrm{x}]$
Take, $1_1=\lim _{\mathrm{x} \rightarrow 2^{-}} \mathrm{f}(\mathrm{x})$
$$
\begin{aligned}
& =\lim _{x \rightarrow 2^{-}}(2 x-[2 x]) \\
& =\lim _{h \rightarrow 0}(2(2-h)-[2(2-h)]) \\
& =\lim _{h \rightarrow 0}(4-2 h-[4-2 h]) \\
& =\lim _{x \rightarrow 0}(4-2 h)-\lim _{h \rightarrow 0}(4-2 h)
\end{aligned}
$$
Greatest integers [ $4-2 \mathrm{~h}]$ lies between
$$
\begin{aligned}
& 3 \leq x \leq 4 \text { then } \lim _{x \rightarrow 0}(4-2 h)=3 \\
& l_1=(4-0-3)=1
\end{aligned} .
$$
$$
\begin{aligned}
& \text { Take } 1_2=\lim _{x \rightarrow 2^{+}} f(x) \\
& =\lim _{x \rightarrow 2^{+}}(2 x-[2 x])
\end{aligned}
$$
$$
\begin{aligned}
& \lim _{h \rightarrow 0}(2(2+h)-[2(2+h)]) \\
& \lim _{h \rightarrow 0}(4+2 h-[4+2 h])
\end{aligned}
$$
$4 \leq x < 5$ then $\lim _{x \rightarrow 0}[4+2 h]=4$
$\{$ here $[4+2 \mathrm{~h}]$ lies between
$4 \leq x < 5$ then $\lim _{x \rightarrow 0}[4+2 h]=4$
$$
1_2=(4-4)=0
$$
So, $1_1+1_2=1+0=1$
Therefore, option (a) is correct.
Take, $1_1=\lim _{\mathrm{x} \rightarrow 2^{-}} \mathrm{f}(\mathrm{x})$
$$
\begin{aligned}
& =\lim _{x \rightarrow 2^{-}}(2 x-[2 x]) \\
& =\lim _{h \rightarrow 0}(2(2-h)-[2(2-h)]) \\
& =\lim _{h \rightarrow 0}(4-2 h-[4-2 h]) \\
& =\lim _{x \rightarrow 0}(4-2 h)-\lim _{h \rightarrow 0}(4-2 h)
\end{aligned}
$$
Greatest integers [ $4-2 \mathrm{~h}]$ lies between
$$
\begin{aligned}
& 3 \leq x \leq 4 \text { then } \lim _{x \rightarrow 0}(4-2 h)=3 \\
& l_1=(4-0-3)=1
\end{aligned} .
$$
$$
\begin{aligned}
& \text { Take } 1_2=\lim _{x \rightarrow 2^{+}} f(x) \\
& =\lim _{x \rightarrow 2^{+}}(2 x-[2 x])
\end{aligned}
$$
$$
\begin{aligned}
& \lim _{h \rightarrow 0}(2(2+h)-[2(2+h)]) \\
& \lim _{h \rightarrow 0}(4+2 h-[4+2 h])
\end{aligned}
$$
$4 \leq x < 5$ then $\lim _{x \rightarrow 0}[4+2 h]=4$
$\{$ here $[4+2 \mathrm{~h}]$ lies between
$4 \leq x < 5$ then $\lim _{x \rightarrow 0}[4+2 h]=4$
$$
1_2=(4-4)=0
$$
So, $1_1+1_2=1+0=1$
Therefore, option (a) is correct.
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