Let [x] denote the greatest integer less than | vedclass

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(A) Given $f(x) = 2x - [2x]$.To find $l_1 = \lim_{x \rightarrow 2^{-}} f(x)$:Let $x = 2 - h$,where $h \rightarrow 0$ and $h > 0$.$l_1 = \lim_{h \right

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(A) Given $f(x) = 2x - [2x]$.To find $l_1 = \lim_{x \rightarrow 2^{-}} f(x)$:Let $x = 2 - h$,where $h \rightarrow 0$ and $h > 0$.$l_1 = \lim_{h \rightarrow 0} (2(2 - h) - [2(2 - h)]) = \lim_{h \rightarrow 0} (4 - 2h - [4 - 2h])$.Since $h$ is a very small positive number,$4 - 2h$ is slightly less than $4$,so $[4 - 2h] = 3$.Thus,$l_1 = 4 - 3 = 1$.To find $l_2 = \lim_{x \rightarrow 2^{+}} f(x)$:Let $x = 2 + h$,where $h \rightarrow 0$ and $h > 0$.$l_2 = \lim_{h \rightarrow 0} (2(2 + h) - [2(2 + h)]) = \lim_{h \rightarrow 0} (4 + 2h - [4 + 2h])$.Since $h$ is a very small positive number,$4 + 2h$ is slightly greater than $4$,so $[4 + 2h] = 4$.Thus,$l_2 = 4 - 4 = 0$.Therefore,$l_1 + l_2 = 1 + 0 = 1$.

Let $[x]$ denote the greatest integer less than or equal to $x$ and $f(x) = 2x - [2x]$. 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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