f(x) = begin{cases} [x^2] - [-x^2], & x neq 3 | vedclass

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(D) For $f(x)$ to be continuous at $x = 3$,we must have $\lim_{x 	o 3} f(x) = f(3) = k$. First,evaluate the limit $\lim_{x 	o 3} ([x^2] - [-x^2])$.

Vedclass · Accepted Answer

No choice of $k$ makes $f(x)$ continuous at $x = 3$

Vedclass · Answer

(D) For $f(x)$ to be continuous at $x = 3$,we must have $\lim_{x 	o 3} f(x) = f(3) = k$. First,evaluate the limit $\lim_{x 	o 3} ([x^2] - [-x^2])$. Recall the property of the greatest integer function: $[y] + [-y] = 0$ if $y \in \mathbb{Z}$,and $[y] + [-y] = -1$ if $y 
otin \mathbb{Z}$. As $x 	o 3$,$x^2 	o 9$. Since $9$ is an integer,we check the left-hand limit $(x 	o 3^-)$ and right-hand limit $(x 	o 3^+)$. For $x 	o 3^-$,$x^2  0$ is very small. Then $[x^2] = 8$ and $[-x^2] = [-9 + h] = -9$. Thus,$\lim_{x 	o 3^-} ([x^2] - [-x^2]) = 8 - (-9) = 17$. For $x 	o 3^+$,$x^2 > 9$,so $x^2 = 9 + h$ where $h > 0$ is very small. Then $[x^2] = 9$ and $[-x^2] = [-9 - h] = -10$. Thus,$\lim_{x 	o 3^+} ([x^2] - [-x^2]) = 9 - (-10) = 19$. Since the left-hand limit $(17)$ is not equal to the right-hand limit $(19)$,the limit $\lim_{x 	o 3} f(x)$ does not exist. Therefore,there is no value of $k$ that makes the function continuous at $x = 3$.

$f(x) = \begin{cases} [x^2] - [-x^2], & x \neq 3 \\ k, & x = 3 \end{cases}$ is continuous at $x = 3$,then $k = $ where $[\cdot]$ is the greatest integer function.

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