Finding limits algebraically - direct substitution

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  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.