Finding limits algebraically - direct substitution

Everything You Need in One Place

Homework problems? Exam preparation? Trying to grasp a concept or just brushing up the basics? Our extensive help & practice library have got you covered.

Learn and Practice With Ease

Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals.

Instant and Unlimited Help

Our personalized learning platform enables you to instantly find the exact walkthrough to your specific type of question. Activate unlimited help now!

  1. No more finding limits "graphically"; Now, finding limits "algebraically"!
    What is Direct Substitution?
  1. Evaluate the limit:
    1. limx3(5x220x+17)\lim_{x \to 3} (5x^2-20x+17)
    2. limx2x3+3x2154x\lim_{x \to -2} \frac{{{x^3} + 3{x^2} - 1}}{{5 - 4x}}
    3. limx0x\lim_{x \to 0}\left| x \right|
    4. limxπ2  sinx2cosx\lim_{x \to \frac{\pi }{2}} \;\frac{{\sin x}}{{2 - \cos x}}
  2. Evaluate the one-sided limit:
    1. limx3(5x220x+17)\lim_{x \to 3^-} (5x^2-20x+17)
      limx3+(5x220x+17)\lim_{x \to 3^+} (5x^2-20x+17)
    2. limx4x4\lim_{x \to {4^ - }} \sqrt {x - 4}
      limx4+x4\lim_{x \to {4^ + }} \sqrt {x - 4}

  3. Finding limits algebraically using direct substitution

    1. Finding limits of a function algebraically by direct substitution
      1. limx1g(x)\lim_{x \to {-1^ - }} g(x)
        limx1+g(x)\lim_{x \to {-1^ + }} g(x)
        limx1g(x)\lim_{x \to {-1}} g(x)
      2. limx4g(x)\lim_{x \to {4^ - }} g(x)
        limx4+g(x)\lim_{x \to {4^ + }} g(x)
        limx4g(x)\lim_{x \to {4}} g(x)
    Topic Notes
    Graphically finding the limit of a function is not always easy, as an alternative, we now shift our focus to finding the limit of a function algebraically. In this section, we will learn how to apply direct substitution to evaluate the limit of a function.
    • if: a function ff is continuous at a number aa
    then: direct substitution can be applied: limxaf(x)=limxa+f(x)=limxaf(x)=f(a)\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) =\lim_{x \to a} f(x)= f(a)

    • Polynomial functions are continuous everywhere, therefore "direct substitution" can ALWAYS be applied to evaluate limits at any number.