exercises 2.2 the limit of a function\nscore: 13.8/17 answered: 15/17\n× question 15\ntextbook videos…

exercises 2.2 the limit of a function\nscore: 13.8/17 answered: 15/17\n× question 15\ntextbook videos +\nscore on last try: 0 of 1 pts. see details for more.\nat least one scored part is incorrect. jump to first changeable incorrect p\n> next question get a similar question you can retry this que\nlet $f(x)=\begin{cases}10 - x - x^{2}&\text{if }xleq3\\2x - 8&\text{if }x > 3end{cases}$\ncalculate the following limits. enter \dne\ if the limit does not exist.\n$lim_{x\rightarrow3^{-}}f(x)=-12$\n$lim_{x\rightarrow3^{+}}f(x)=4$\n$lim_{x\rightarrow3}f(x)=dne$\nquestion help: video written example message instructor
Answer
Explanation:
Step1: Calculate left - hand limit
For $x\to3^-$, use $f(x)=10 - x - x^{2}$. Substitute $x = 3$ into $10 - x - x^{2}$: $10-3 - 3^{2}=10 - 3-9=-2$.
Step2: Calculate right - hand limit
For $x\to3^+$, use $f(x)=2x - 8$. Substitute $x = 3$ into $2x - 8$: $2\times3 - 8=6 - 8=-2$.
Step3: Determine the overall limit
Since $\lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x)=-2$, then $\lim_{x\to3}f(x)=-2$.
Answer:
$\lim_{x\to3^-}f(x)=-2$ $\lim_{x\to3^+}f(x)=-2$ $\lim_{x\to3}f(x)=-2$