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

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(A) For $f(x)$ to be continuous at $x = 2,$ the condition $\lim_{x 	o 2} f(x) = f(2)$ must hold. First,we evaluate the limit $\lim_{x 	o 2} ([x] + [

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(A) For $f(x)$ to be continuous at $x = 2,$ the condition $\lim_{x 	o 2} f(x) = f(2)$ must hold. First,we evaluate the limit $\lim_{x 	o 2} ([x] + [-x])$. We know that for any $x 
otin \mathbb{Z},$ $[x] + [-x] = -1.$ Since the limit $x 	o 2$ considers values of $x$ arbitrarily close to $2$ but not equal to $2,$ we have: $\lim_{x 	o 2} ([x] + [-x]) = -1.$ For continuity,we require $f(2) = \lim_{x 	o 2} f(x).$ Given $f(2) = \lambda,$ we have $\lambda = -1.$

If $f(x) = \begin{cases} [x] + [-x], & x \ne 2 \\ \lambda, & x = 2 \end{cases},$ then $f$ is continuous at $x = 2,$ provided $\lambda$ is (where $[.]$ is the Greatest Integer Function).

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