AP Calculus AB Unit 1 Review

1. Hole/Point discontinuity (REMOVABLE)

2. Jump (NON-REMOVABLE)

3. Vertical Asymptote (NON-REMOVABLE)

If $$f\left(x\right)$$ is continuous on $$\left[a,b\right]$$ and $$y$$ is between $$f\left(a\right)$$ and $$f\left(b\right)$$ , then there must be at least one $$c$$ in $$\left(a,b\right)$$ such that $$f\left(c\right)=y$$ .

Since $$\lim_{x\rightarrow2}g\left(x\right)=\frac{14-26}{2-5}=\frac{-12}{-3}=4$$ and $$\lim_{x\rightarrow2}h\left(x\right)=\frac{6+14}{4+1}=\frac{20}{5}=4$$ , it can be concluded from the squeeze theorem that $$\lim_{x\rightarrow2}f\left(x\right)=4$$ because $$g\left(x\right)\le f\left(x\right)\le h\left(x\right)$$ for $$0<x<5$$ .

Since $$\lim_{x\rightarrow6}f\left(x\right)<5<10=f\left(6\right)$$ , it is false that $$\lim_{x\rightarrow6}f\left(x\right)=f\left(6\right)$$ . Therefore $$f$$ cannot be continuous at $$x=6$$ .

$$\lim_{x\rightarrow2^+}f\left(x\right)$$ is the limit of $$f\left(x\right)$$ as $$x$$ approaches $$2$$ from the right. The portion of the graph to the right of $$x=2$$ has y-values approaching $$1$$ as the x-values approach $$2$$ .

$$\lim_{x\rightarrow2^+}f\left(x\right)$$ is the limit of $$f\left(x\right)$$ as $$x$$ approaches $$2$$ from the right. The portion of the graph to the right of $$x=2$$ has y-values approaching $$3$$ as the x-values approach $$2$$ .

$$f\left(x\right)=\frac{x^2-4}{x^2+x-6}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x+2}{x+3}$$ for $$x$$ close to, but not equal to, $$2$$ . Therefore, $$\lim_{x\rightarrow2}f\left(x\right)=\lim_{x\rightarrow2}\frac{x+2}{x+3}=\frac{2+2}{2+3}=\frac{4}{5}$$ .

Continuous because the graph can be drawn without lifting your pen.

$$f\left(x\right)=\frac{\frac{1}{x}-1}{x-1}=\frac{\frac{1}{x}-\frac{x}{x}}{x-1}=\frac{1-x}{x\left(x-1\right)}=\frac{-1}{x}$$ for $$x$$ close to, but not equal to, $$1$$ . Therefore $$\lim_{x\rightarrow1}f\left(x\right)=\lim_{x\rightarrow1}\left(-\frac{1}{x}\right)$$ .

This graph has a jump discontinuity at $$x=3$$ , because $$\lim_{x\rightarrow3^-}f\left(x\right)$$ and $$\lim_{x\rightarrow3^+}f\left(x\right)$$ are finite but not equal to each other.