Practice Exam B Answer Key

Left $2$, right $2$, two-sided limit $2$, function value $5$; not continuous, removable discontinuity.

Cancel $x-2$: $(x^2+2x+4)/(x+2)\to12/4=3$.

Rationalization gives $5/[\sqrt{1+4x}+\sqrt{1-x}]\to5/2$.

$9\cdot(1/2)=9/2$.

$2/5$.

Squeeze gives $0$.

Hole at $1$; vertical asymptote $x=3$. Both one-sided limits at $3$ are $+\infty$.

$+\infty$.

Division gives polynomial asymptote $y=2x$.

$-5$.

Continuity at $2$ gives $6=6+a$, so $a=0$; continuity at $1$ then gives $a+b=3$, so $b=3$.

The polynomial is continuous, $f(0)=1$, $f(1)=-1$. First midpoint $0.5$ is negative, so interval $(0,0.5)$. Second midpoint $0.25$ is positive, so interval $(0.25,0.5)$.

Choose $\delta=\varepsilon/4$; then $|(4x-3)-1|=4|x-1|<4\delta=\varepsilon$.

Counterexample: $f(x)=1/x$ on $[-1,1]$. Endpoint signs differ, but the function is not continuous on the interval. The missing hypothesis is continuity.