1. Find \(\lim_{x\rightarrow 3} -x^3 + 2x^2 +2\) by direct substitution.
x <- 3
(poly <- -1 * x^3 + 2*x^2 +2)
## [1] -7

\(\lim_{x\rightarrow 3} = -7\)

  1. Find \(\lim_{x\rightarrow -1} -x^2 + 2x +1\) by direct substitution.
x <- -1
(poly <- -1 * x^2 + 2*x +1)
## [1] -2

\(\lim_{x\rightarrow -1} = -2\)

  1. Find \(\lim_{x\rightarrow \frac{-\pi}{4}} - \cot x\) by direct substitution.
x <- -1 * pi/4
(poly <- -1/tan(x))
## [1] 1

\(\lim_{x\rightarrow \frac{-\pi}{4}} = 1\)

  1. Find \(\lim_{x\rightarrow 0} 2 \cos (2x)\) by direct substitution.
x <- 0
(poly <- 2*cos(2*x))
## [1] 2

\(\lim_{x\rightarrow 0} = 2\)

  1. Find \(\lim_{x\rightarrow \infty} \dfrac{-x+2}{x^2 + x + 1}\) by sketching the graph of \(f(x) = \dfrac{-x+2}{x^2 + x + 1}\).
f_x <- function(x){
  (-x+2)/(x^2+x+1)}
curve(f_x, -5, 5)
abline(h=0)

\(\lim_{x\rightarrow \infty} = 0\)

  1. Find \(\lim_{x\rightarrow \infty} \dfrac{3x^3}{2x^2 +3}\) by sketching the graph of \(f(x) = \dfrac{3x^3}{2x^2 +3}\).
f_x <- function(x){
  (3*x^3)/(2*x^2+3)}
curve(f_x, -5, 5)
abline(h=0)

\(\lim_{x\rightarrow \infty} = DNE\)

  1. Find \(\lim_{x\rightarrow 2} f(x)\) \[f(x) = \begin{cases} -5, & x \le 2\\ -x^2 - 1, & x > 2 \end{cases}\]

lh_lim <- -5 rh_lim <- -1*2^2-1 lh_lim; rh_lim; lh_lim==rh_lim

\(\lim_{x\rightarrow 2} = -5\)

  1. Find \(\lim_{x\rightarrow 0} f(x)\) \[f(x) = \begin{cases} -\frac{x}{2} + \frac{3}{2}, & x < 0 \\ -2x , & x >= 0 \end{cases}\]
lh_lim <- (0/2) + (3/2)
rh_lim <- -2*0
lh_lim; rh_lim; lh_lim==rh_lim
## [1] 1.5
## [1] 0
## [1] FALSE

\(\lim_{x\rightarrow 2} = DNE\)

Note: We did not do right or left hand limits this year so problems like 9 & 10 will not be on the test.

  1. Find \(\lim_{x\rightarrow 1^+} f(x)\) \[f(x) = \begin{cases} 3, & x < 1\\ x^2 + 2x, & x \ge 1 \end{cases}\]
rh_lim <- 1^2 + 2*1
rh_lim
## [1] 3

\(\lim_{x\rightarrow 1} = 3\)

Note: If you treated this as a two sided limit, the type we learned. The limit would be 3 also.

  1. Find \(\lim_{x\rightarrow -1^-} f(x)\) \[f(x) = \begin{cases} \frac{x}{2} + \frac{1}{2}, & x < -1\\ -1, & x \ge -1 \end{cases}\]
lh_lim <- -1/2 + 1/2
lh_lim
## [1] 0

\(\lim_{x\rightarrow 1} = 0\)

Note: If you treated this as a two sided limit, the type we learned. The answer would be DNE.

Evaluate the following algebraically using the method taught in class

  1. \[\lim_{x\rightarrow - \infty} \dfrac{x^4}{4x^2-1}= DNE\]

  2. \[\lim_{x\rightarrow \infty} \dfrac{2x}{x+2}= 2\]

  3. \[\lim_{x\rightarrow \infty} \dfrac{x^2}{x^2+9}= 1\]

  4. \[\lim_{x\rightarrow \infty} -\dfrac{x^2}{x^2+9}= -1\]

Solve each inequaltiy, use a number line in your work.

  1. \(0<x^2+8x+12\)
curve(x^2+8*x+12, -8, 0)
abline(h=0)

\((-\infty, -2) \cup (-6, \infty)\)

  1. \(0 \ge 3x^2 - 18x +24\)
curve(3*x^2 - 18*x + 24, -5, 5)
abline(h=0)

\([2, 4]\)

  1. \(x^2(x-3)(x+4) \ge 0\)
curve(x^2*(x-3)*(x+4), -5, 5)
abline(h=0)

\((-\infty, -4] \cup [0,0] \cup [3, \infty)\)

  1. \(\dfrac{x-3}{x+4}> 0\)
curve((x-3)/(x+4), -10, 10, ylim=c(-10,10))
abline(h=0)

\((-\infty, -4)\cup (3,\infty)\)

Find the horizontal asymptote by setting up a limit and solving it with the official method.

  1. \(f(x)=\dfrac{2x^3}{4x^2-1}\)

\[\lim_{x\rightarrow \infty}\frac{2x^3}{4x^2-1}=DNE\]

  1. \(f(x)=\dfrac{x}{x^2+2x+2}\)

\[\lim_{x\rightarrow \infty}\frac{x}{x^2+2x+2}=0\]

  1. Find the partial fraction decomposition of \(\dfrac{2}{(x-1)(x+1)}\)

\[\frac{2}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}\] \[A(x+1) + B(x-1) = 2\] \[(A+B)x + (A-B) = 0x+2\]

This becomes: \[\begin{cases} A+B=0\\A - B=2\end{cases}\] \(A=1\) and \(B=-1\) so \[\frac{2}{(x-1)(x+1)} = \frac{1}{x-1} - \frac{1}{x+1}\]

  1. Find the partial fraction decomposition of \(\dfrac{24}{x^2-4}\)

\[\frac{24}{x^2-4} = \frac{A}{x-2} + \frac{B}{x+2}\] \[A(x+2) + B(x-2) = 24\] \[(A+B)x + (2A-2B) = 0x+24\]

This becomes: \[\begin{cases} A+B=0\\2A-2B=24\end{cases}\] \(A=6\) and \(B=-6\) so \[\frac{24}{x^2-4} = \frac{6}{x-2} - \frac{6}{x+2}\]

  1. Find the partial fraction decomposition of \(\dfrac{x^3+47}{x^3+x^2}\)

\[\frac{x^3+47}{x^3+x^2} = 1+\frac{-x^2+47}{x^3+x^2} = 1 + \frac{A}{x^2}+\frac{B}{x}+\frac{C}{x+1}\] \[(x^3+x^2) + A(x+1) + Bx(x+1) + Cx^2= x^3+47\] \[x^3 +(1+B+C)x^2+(A+B)x + 47 = x^3+0x^2+0x+47\]

This becomes: \[\begin{cases} x^3=x^3\\1+B+C=0\\A+B=0\\A=47\end{cases}\] \(A=47\), \(B=-47\), and \(C=46\) so \[\frac{x^3+47}{x^3+x^2} = 1 + \frac{47}{x^2}-\frac{47}{x}+\frac{46}{x+1}\]

Find \[\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\] where…

  1. \(f(x)=-2x^2+2\)

\[\lim_{h \rightarrow 0} =-4x\]

  1. \(f(x)=-4x+1\)

\[\lim_{h \rightarrow 0} =-4\]

  1. \(f(x)=-5x^2-3\) \[\lim_{h \rightarrow 0} =-10x\]

  2. Graph \[x=y^2-8y+11\]

  1. Graph \[x=-2y^2+20y-49\]
  1. Graph \[x=y^3-3y^2+2\]
  1. Graph \[x=y^4+2y^2\]
  1. The answers are: