Finding limits algebraically - direct substitution

In our previous courses in Mathematics, we learned about the different concepts about function. This time, in this chapter, we will talk about the limits of function. In mathematics, a limit is defined as the value that a function or sequence "approaches" given a certain value of x.

Suppose we are given the function f(x) (x21)(x1)\frac{(x^2-1)}{(x-1)}, if we designate 1 as the value of x, we know that as the result would be 0/0. This is an indeterminate value, so we need to come up with a limit. We know that the function will be indeterminate is x=1, but if we try approaching one closer and closer, we will see that the solution is close to 2, so we say that the limit of f(x) (x21)(x1)\frac{(x^2-1)}{(x-1)}, as x approaches 1 is 2.

Learning about limits is a very important part of mathematical analysis because it defines the continuity of a function. It is also used in deriving, and integrating an expression.

In this chapter, we will focus our discussion about limits. This chapter will have eight parts. For the first part, we will be finding limits graphically. We will look into the definition of the right hand limit and the left hand limit. By knowing the definition of these two limits, we will learn the different limits from both sides. The definition of these two limits can be summarized by this expression:

limxaf(x)=L\lim_{x \to a} f(x) = L if and only if limxa+f(x)=L\lim_{x \to a^+} f(x) = L and limxaf(x)=L\lim_{x \to a^-} f(x) = L

For the second part of this chapter, we will learn about Continuity. A function is said to be continuous if only and if limxaf(x)=limxa+f(x)=f(a)\lim_{x \to a^-} f(x)=\lim_{x \to a^+} f(x)=f(a). We will get to see how the graph of a continuous function looks like.

For the third and fourth part of this chapter, we will be finding limits algebraically through direct substitution and through other methods if direct substitution is not applicable. In some cases direct substitution is not applicable because there is a hole in the function, so we will find the limits by conjugates multiplication.

For the next part of the chapter, we will learn about infinite limits and the vertical asymptotes. We will get to learn about the limits at infinity and the horizontal asymptotes in the following part. We will also look at the end behaviors of a limit as it approaches infinity.

Finally, for the last parts of this chapter, we will discuss about the intermediate value theorem and the squeeze theorem.

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.


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)

      - 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)
Teacher pug

Finding limits algebraically - direct substitution

Don't just watch, practice makes perfect.

We have over 570 practice questions in Calculus 1 for you to master.