explain why the function is discontinuous at the given number a. (select all that apply.)\n…

explain why the function is discontinuous at the given number a. (select all that apply.)\n f(x)=left{\begin{array}{ll}cos (x) & \text { if } x<0 \\0 & \text { if } x=0 \\1-x^{2} & \text { if } x>0end{array}\right. quad a = 0 \n(square lim _{x \rightarrow 0} f(x)) does not exist.\n(square f(0)) and (lim _{x \rightarrow 0} f(x)) are finite, but are not equal.\n(square lim _{x \rightarrow 0^{+}} f(x)) and (lim _{x \rightarrow 0^{-}} f(x)) are finite, but are not equal.\n(square f(0)) is undefined.\n(square) none of the above\nsketch the graph of the function.

explain why the function is discontinuous at the given number a. (select all that apply.)\n f(x)=left{\begin{array}{ll}cos (x) & \text { if } x<0 \\0 & \text { if } x=0 \\1-x^{2} & \text { if } x>0end{array}\right. quad a = 0 \n(square lim _{x \rightarrow 0} f(x)) does not exist.\n(square f(0)) and (lim _{x \rightarrow 0} f(x)) are finite, but are not equal.\n(square lim _{x \rightarrow 0^{+}} f(x)) and (lim _{x \rightarrow 0^{-}} f(x)) are finite, but are not equal.\n(square f(0)) is undefined.\n(square) none of the above\nsketch the graph of the function.

Answer

Explanation:

Step1: Calculate the left - hand limit

For (x\to0^{-}), (f(x)=\cos(x)). Then (\lim_{x\to0^{-}}f(x)=\lim_{x\to0^{-}}\cos(x)=\cos(0) = 1)

Step2: Calculate the right - hand limit

For (x\to0^{+}), (f(x)=1 - x^{2}). Then (\lim_{x\to0^{+}}f(x)=\lim_{x\to0^{+}}(1 - x^{2})=1-0^{2}=1)

Step3: Calculate the limit as (x\to0)

Since (\lim_{x\to0^{-}}f(x)=\lim_{x\to0^{+}}f(x) = 1), then (\lim_{x\to0}f(x)=1)

Step4: Evaluate (f(0))

Given (f(0) = 0)

Answer:

(\lim_{x\to0}f(x)=1) and (f(0)=0). So (f(0)) and (\lim_{x\to0}f(x)) are finite, but are not equal.