MATH 1830

Unit 1 Limits

1.2 Finding Limits Algebraically/Infinte Limits/Continuity



1.2A Video

Limits: An Algebraic Approach

Find each indicated quantity, if it exists.

  1. $\mathop {\lim }\limits_{x \to - 5} \;2{x^2} + 10x + 7 =$

    $2{\left( { - 5} \right)^2} + 10\left( { - 5} \right) + 7 =$

    $7$

  2. $f\left( x \right) = \;\frac{{3{x^2}\; + 2x - 1}}{{x{\;^2} + 3x + 2}}$

    1. $\underset{x\rightarrow -3}{lim}f(x)=$

      $\underset{x\rightarrow -3}{lim}f(x)=\frac{3{{\left( -3 \right)}^{2}}~+2\left( -3 \right)-1}{\left( -3 \right){{~}^{2}}+3\left( -3 \right)+2}=\frac{20}{2}$

      $10$

    2. $\underset{x\rightarrow -1}{lim}f(x)=$

      $\frac{3{{\left( -1 \right)}^{2}}~+2\left( -1 \right)-1}{\left( -1 \right){{~}^{2}}+3\left( -1 \right)+2}=\frac{0}{0}$

      Indeterminate form. Factor, Reduce, Try Again.

      $\underset{x\rightarrow -1}{lim}f(x)=\underset{x\rightarrow -1}{lim}\frac{\left( 3x-1 \right)\left( x+1 \right)}{\left( x+1 \right)\left( x+2 \right)}=\underset{x\rightarrow -1}{lim}\frac{\left( 3x-1 \right)}{\left( x+2 \right)}=\frac{-4}{1}$

      $-4$

    3. $\mathop {\lim }\limits_{x\; \to \; - 2} f\left( x \right)$ $=\frac{7}{0}\,$ Does not exist


    3. 1.2B Video

    Find the horizontal asymptotes and where the function is discontinuous.

  3. $f(x)=\frac{2{{x}^{2}}-7x-15}{{{x}^{2}}+3x-40}$

    $f(x)=\frac{(2x+3)(x-5)}{(x+8)(x-5)}$

    HA: $y=2$

    VA: $x=-8$

    Hole: $x=5$

    Discontinuous at x=-8 and x=5



    4. 1.2 Lecture

  4. Find the indicated quantity, if it exists.

    1. $\underset{x\rightarrow 0}{lim} \frac{{{x^2}\; - 15x + 50}}{{{{\left( {x - 10} \right)}^2}}}$

      $=\frac{50}{100}=\frac12$

    2. $\underset{x\rightarrow 5}{lim} \frac{{{x^2}\; - 15x + 50}}{{{{\left( {x - 10} \right)}^2}}}$

      $=\frac{0}{25}=0$

    3. $\underset{x\rightarrow 10}{lim} \frac{{{x^2}\; - 15x + 50}}{{{{\left( {x - 10} \right)}^2}}}$

      $=\frac{0}{0}$

      Indeterminate form. Factor, reduce, try again.

      $\underset{x\rightarrow 10}{lim}\,\frac{\left( x-5 \right)\left( x-10 \right)}{{{\left( x-10 \right)}^{2}}}=\underset{x\rightarrow 10}{lim}\,\frac{\left( x-5 \right)}{\left( x-10 \right)}=\frac{5}{0}$

      The limit does not exist. Vertical asymptote at x=10. On inspection of the graph, the limit as x approaches 10 from the left and the limit as x approaches 10 from the right are not equal.

    4. $\underset{x\rightarrow4}{lim}\frac{\sqrt x-2}{x-4}$

      $\underset{x\rightarrow4}{lim}\frac{\sqrt x-2}{x-4}=\frac00$

      Indeterminate form. Technically will factor but hard to do, so graph the function to determine limit.

      $\underset{x\rightarrow4}{lim}\frac{\sqrt x-2}{x-4}=0.25$

  5. Find all the vertical asymptotes, horizontal asymptotes, and holes of the function, showing all your work.
    1. $f\left( x \right) = \;\frac{{2{x^2}\;\; - \;\;32}}{{{x^2}\; + \;5x\; + \;4}}$

      $f\left( x \right) = \;\frac{{2\left( {x + 4} \right)\left( {x - 4} \right)}}{{\left( {x + 4} \right)\left( {x + 1} \right)}}$

      Vertical Asymptote: x=-1

      Hole: x=-4

      Horizontal Asymptote: y=2





    2. Where is this function discontinuous?

      Discontinuous at x=-4 and x=-1

     

    1.2 Group Work

  6. $\mathop {\lim }\limits_{x \to 4} \;{x^2} - 5x + 1 =$

    ${4^2} - 5\left( 4 \right) + 1 =$

    $- 3$

  7. $\mathop {\lim }\limits_{x \to - 5} \;\frac{{x^2\; + 7x + 10}}{{x{\;^2} + 2x -15}}$

    $\frac{(-5)^2+7(-5)+10}{(-5)^2+2(-5)-15}=\frac{0}{0}$

    Indeterminate form. Factor, Reduce, Try Again.

    $\underset{x\rightarrow -5}{lim}f(x)=\underset{x\rightarrow -5}{lim}\frac{\left( x+5 \right)\left( x+2 \right)}{\left( x+5 \right)\left( x-3 \right)}=\underset{x\rightarrow -5}{lim}\frac{\left( x+2 \right)}{\left( x-3 \right)}=\frac{3}{8}$

    $\frac{3}{8}$

  8. $\mathop {\lim }\limits_{x \to 4} \;\frac{{x^2\; -16}}{{3x{\;^2} -13x +4}}$

    $\frac{(4)^2-16}{3(4)^2-13(4)+4}=\frac{0}{0}$

    Indeterminate form. Factor, Reduce, Try Again.

    $\underset{x\rightarrow 4}{lim}f(x)=\underset{x\rightarrow 4}{lim}\frac{\left( x+4 \right)\left( x-4 \right)}{\left( 3x-1 \right)\left( x-4 \right)}=\underset{x\rightarrow 4}{lim}\frac{\left( x+4 \right)}{\left( 3x-1 \right)}=\frac{8}{11}$

    $\frac{8}{11}$

    9. Find all the vertical asymptotes, horizontal asymptotes, and holes of the function. Also identify where the function is discontinous.

  9. $f\left( x \right)\; = \frac{{{x^2}\;\; - \;\;9}}{{{x^2}\;\; - \;\;4}}$

    $f\left( x \right) = \;\frac{{\left( {x + 3} \right)\left( {x - 3} \right)}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}$

    Vertical Asymptote: x=2 and x=-2

    Hole: none

    Horizontal Asymptote: y=1

    Discontinuous at x=2 and x=-2

  10. $f\left( x \right)\; = \frac{{x\;\; + 2}}{{{x^2}\; + \;3}}$

    Vertical Asymptote: none

    Hole: none

    Horizontal Asymptote: y=0

    This function is continuous.

  11. $f\left( x \right)\; = \frac{{{x^2}\; - 3x - \;10}}{{{x^2}\; - \;4x - 5}}$

    $f(x) = \frac{{\left( {x - 5} \right)\left( {x + 2} \right)}}{{\left( {x - 5} \right)\left( {x + 1} \right)}}$

    Vertical Asymptote: x=-1

    Hole: x=5

    Horizontal Asymptote: y=1

    Discontinuous at x=-1 and x=5

  12. $f(x) = \frac{{{x^2} + 5x - 14}}{{x - 2}}$

    Vertical Asymptote: none

    Hole: x=2

    Horizontal Asymptote: none

    Discontinuous at x=2

    13.  

    1.2 Additional Practice

    Compute the limits. If the limit does not exist, explain why.

  13. $\underset{x\rightarrow3}{lim}\,\ \frac{{{x}^{2}}+x-12}{x-3}\ = $

    $\frac{{{3}^{2}}+3-12}{3-3}=\frac{0}{0}$

    Indeterminant form: Factor, reduce, take the limit again

    $\underset{x\rightarrow3}{lim}\,\ \frac{(x+4)(x-3)}{x-3}$

    $=\underset{x\rightarrow3}{lim}\,\ \left( x+4 \right)$

    $=3+4=$

    $7$

  14. $\underset{x\rightarrow1}{lim}\,\ \frac{{{x}^{2}}+x-12}{x-3}\ = $

    $\frac{{{1}^{2}}+1-12}{1-3}=\frac{-10}{-2}=$

    $5$

  15. $\underset{x\rightarrow10}{lim}\,\ 10 =$

    $10$

  16. $\underset{x\rightarrow4}{lim}\,\ 3{{x}^{2}}-5x = $

    $3{{(4)}^{2}}-5(4)$

    $=3(16)-20$

    $=48-20=$

    $28$

  17. $\underset{x\rightarrow0}{lim}\,\ \frac{4x-5{{x}^{2}}}{x-1} = $

    $\frac{4(0)-5{{(0)}^{2}}}{0-1}=\frac{0}{-1}=$

    $0$

  18. $\underset{x\rightarrow1}{lim}\,\ \frac{{{x}^{2}}-1}{x-1} = $

    $\frac{{{1}^{2}}-1}{1-1}=\frac{0}{0}$

    Indeterminant form: Factor, reduce, take the limit again

    $\underset{x\rightarrow1}{lim}\frac{\left(x+1\right)\left(x-1\right)}{x-1}$

    $=\,\underset{x\rightarrow1}{lim}\,\left( x+1 \right)$

    $=1+1$

    $2$

  19. $\underset{x\rightarrow0^{+}}{lim}\,\ \frac{\sqrt{2-{{x}^{2}}}}{x} = $

    $\frac{\sqrt{2-{{0}^{2}}}}{0}=$

    $\frac{\sqrt{2}}{0}$ , which is undefined

  20. $\underset{x\rightarrow0^{+}}{lim}\,\ \frac{\sqrt{2-{{x}^{2}}}}{x+1} =$

    $\frac{\sqrt{2-{{0}^{2}}}}{0+1}=\frac{\sqrt{2}}{1}=$

    $\sqrt{2}$ /li>

  21. $\underset{x\rightarrow2}{lim}\,\ {{({{x}^{2}}+4)}^{3}} =$

    ${{({{2}^{2}}+4)}^{3}}={{(4+4)}^{3}}={{(8)}^{3}}=$

    $512$

    http://www.whitman.edu/mathematics/california_calculus/calculus.pdf

     

    Find the vertical and horizontal asymptotes and holes, if they exist. Also identify where the function is discontinous.

  22. $f(x)=\frac{2x-7}{x-4}\ $

    VA: $x=4$

    HA: $y=2$

    Holes: none

    Discontinuous at $x=4$

  23. $f(x)=\ \frac{{{x}^{2}}-5}{{{x}^{3}}+{{x}^{2}}+1}\ $

    VA: $x=-1.47$

    HA: $y=0$

    Holes: none

    Discontinuous at $x=-1.47$

  24. $f(x)=\frac{{{x}^{3}}+6{{x}^{2}}+8x}{{{x}^{2}}-16}\ $

    VA: $x=4$

    HA: none

    Holes: $x=-4$

    Discontinuous at $x=4$ and $x=-4$

  25. $f(x)=\frac{x-5}{\sqrt{4{{x}^{2}}+8}}\ $

    VA: none

    HA: $y=0.5$

    Holes: none

    Continuous

  26. $f(x)=\ \frac{{{x}^{2}}-1}{{{x}^{2}}-x-2}$.

    VA: $x=2$

    HA: $y=1$

    Holes: $x=-1$

    Discontinuous at $x=2$ and $x=-1$

    27.  

    Extra Practice for Limits and Asymptotes

    Find the limits.

    Coordinate Plane: -10 to 10 on x- and y-axes. Vertical Asymptotes at x =-4 and x=2. Three curves graphed on the plane. Left curve begins at top left of graph, crosses the x-axis at approximately -7.75, continues down until approximately (-6,-4), is horizontal until x=-4.5, then increases very close to the vertical asymptote at x=-4. The middle curve rises from the bottom of the graph along the vertical asymptote at x=-4 until the point (-2,-1). The curve then decreases until its endpoint at (2,-3). This endpoint is a solid dot. The curve on the right third of the graph is decreasing from the top of the coordinate plane along the right side of the vertical asymptote at x=2. It crosses the x-axis at approximately x=2.25 and continues decreasing. This curve decreases more slowly (level out) beginning at x=3 to almost a horizontal line at y=-5.

    1. $\underset{x\rightarrow-4^{-}}{lim}\,f(x)=$ $\infty $

    2. $\underset{x\rightarrow-4^{+}}{lim}\,f(x)=$ $-\infty $

    3. $\underset{x\rightarrow-4}{lim}\,f(x)=$ Does Not Exist

    4. $f(-4)=$ undefined

    5. $\underset{x\rightarrow2^{-}}{lim}\,f(x)=$ $-3$

    6. $\underset{x\rightarrow2^{+}}{lim}\,f(x)=$ $\infty $

    7. $\underset{x\rightarrow2}{lim}\,f(x)=$ Does Not Exist

    8. $f(2)=$ $-3$

      Find the limit.

    9. $\underset{x\rightarrow1}{lim}\,\frac{10x}{x-1}=$

      $\frac{10(1)}{1-1}=\frac{10}{0}$

      Does Not Exist

    10. $\underset{x\rightarrow-12}{lim}\,\frac{{{x}^{2}}+11x-12}{x+12}=$

      $\underset{x\rightarrow-12}{lim}\,\frac{(x+12)(x-1)}{(x+12)}=\underset{x\rightarrow-12}{lim}\,(x-1)=$

      $-13$

      Find the vertical and horizontal asymptotes and holes, if they exist.

    11. $f(x)=\frac{9x}{x+7}$

      VA: $x=-7$

      HA: $y=9$

      Holes: none

    12. $f(x)=9{{x}^{8}}+7{{x}^{6}}+21$

      VA: none

      HA: none

      Holes: none

    13. $f(x)=\frac{10{{x}^{2}}+49}{{{x}^{2}}-49}$

      VA: $x=7$ and $x=-7$

      HA: $y=10$

      Holes: none

    14. $f(x)=\frac{6x+7}{7{{x}^{2}}+4}$

      VA: none

      HA: $y=0$

      Holes: none