Finding limits algebraically - direct substitution - Limits

Finding limits algebraically - direct substitution

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.

Lessons

Notes:
if: a function ff is continuous at a number aa

then: direct substitution can be applied: limxaf(x)=limxa+f(x)=f(a)\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)

note: polynomial functions are continuous everywhere, therefore “direct substitution” can always be applied to evaluate limits at any number.
    • a)
      State the value of the limit from the graph of the function f(x)=1x2f\left( x \right) = \frac{1}{{x - 2}}




      - limx3f(x)\lim_{x \to 3} f(x)
      - limx2.5f(x)\lim_{x \to 2.5} f(x)
      - limx0f(x)\lim_{x \to 0} f(x)
    • b)
      Evaluate:

      - f(3)f(3)
      - f(2.5)f(2.5)
      - f(0)f(0)
  • 2.
    Evaluate the limit:
  • 3.
    Evaluate the one-sided limit:
    • a)
      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)
    • b)
      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)
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Finding limits algebraically - direct substitution

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