Functions Question 683
Question: If $ f(x)= \begin{cases} & \frac{x^{2}-1}{x+1},,\text{when }x\ne -1 \\ & -2,,\text{when }x=-1 \\ \end{cases} . $ ,then
Options:
A) $ \underset{x\to {{(-1)}^{-}}}{\mathop{\lim }},f(x)=-2 $
B) $ \underset{x\to {{(-1)}^{+}}}{\mathop{\lim }},f(x)=-2 $
C) $ f(x) $ is continuous at $ x=-1 $
D) All the above are correct
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{x\to 1-}{\mathop{\lim }},f(x)=-2 $ and $ f(-1)=-,2. $
BETA
