question 3 - no calculator\nif $f(x)=\\begin{...
question 3 - no calculator\nif $f(x)=\\begin{cases}e^{2}+e^{x}&\\text{for }0 < x < 2\\x^{2}e^{x}-xe^{x}&\\text{for }2\\leq x < 8\\end{cases}$, then $\\lim_{x\\to2}f(x)$ is
Answer
# Explanation:
## Step1: Find left - hand limit
We find $\lim_{x\rightarrow2^{-}}f(x)$. Since for $0 < x<2$, $f(x)=e^{2}+e^{x}$, then $\lim_{x\rightarrow2^{-}}f(x)=e^{2}+e^{2}=2e^{2}$.
## Step2: Find right - hand limit
We find $\lim_{x\rightarrow2^{+}}f(x)$. Since for $2\leq x < 8$, $f(x)=x^{2}e^{x}-xe^{x}$, then $\lim_{x\rightarrow2^{+}}f(x)=2^{2}e^{2}-2e^{2}=4e^{2}-2e^{2}=2e^{2}$.
## Step3: Determine the limit
Since $\lim_{x\rightarrow2^{-}}f(x)=\lim_{x\rightarrow2^{+}}f(x) = 2e^{2}$, then $\lim_{x\rightarrow2}f(x)=2e^{2}$.
# Answer:
$2e^{2}$