If f(x) = begin{cases} [x] + [-x], & x neq 2 | vedclass

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(A) For the function $f(x)$ to be continuous at $x = 2$,the limit of $f(x)$ as $x 	o 2$ must exist and be equal to $f(2)$.Given $f(x) = [x] + [-x]$ f

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$-1$

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(A) For the function $f(x)$ to be continuous at $x = 2$,the limit of $f(x)$ as $x 	o 2$ must exist and be equal to $f(2)$.Given $f(x) = [x] + [-x]$ for $x 
eq 2$.We know that for any real number $x$,$[x] + [-x] = \begin{cases} 0, & 	ext{if } x \in \mathbb{Z} \ -1, & 	ext{if } x 
otin \mathbb{Z} \end{cases}$.Since $x = 2$ is an integer,we evaluate the limit as $x$ approaches $2$ from the left $(x 	o 2^-)$ and from the right $(x 	o 2^+)$.For $x$ near $2$ but $x 
eq 2$,$x$ is not an integer,so $[x] + [-x] = -1$.Therefore,$\lim_{x 	o 2} f(x) = -1$.For continuity,$f(2) = \lim_{x 	o 2} f(x)$,which implies $\lambda = -1$.

If $f(x) = \begin{cases} [x] + [-x], & x \neq 2 \\ \lambda, & x = 2 \end{cases}$ is continuous at $x = 2$,then $\lambda = $ (where $[.]$ denotes the greatest integer function).

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