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    Math Deficiency – II
    MD-002
    Progress0 / 32 topics
    Topics
    1. Complex Numbers2. Arithmetic with Complex Numbers (Add, subtract, multiply and divide complex numbers)3. Trigonometric Polar Form of Complex Numbers4. De Moivre's Theorem and nth Roots5. Recursion6. Sequences and Series7. Sigma Notation8. Arithmetic Series9. Geometric Series (Sum infinite and finite geometric series and categorize geometric series)10. Counting with Permutations and Combinations11. Basic Probability12. Binomial Theorem13. Limit: Notation, Graphs to Find Limits, Tables to Find Limits14. Substitution to Find Limits, Rationalization to Find Limits15. One Sided Limits and Continuity16. Rate of Change: Instantaneous Rate of Change17. Tangent Lines and Rates of Change18. Derivatives: The Derivative Function19. Introduction to Techniques of Differentiation20. The Product and Quotient Rules21. Derivatives of Trigonometric Functions22. The Chain Rule23. Derivatives of Logarithmic Functions24. Derivatives of Exponential and Inverse Trigonometric Functions25. Increase, Decrease, and Concavity26. Relative Extrema, Absolute Maxima and Minima27. Integrals: An Overview of the Area Problem28. Area Under a Curve29. The Indefinite Integral30. Integration by Substitution31. The Definition of Area as a Limit; Sigma Notation32. The Definite Integral
    MD-002›Substitution to Find Limits, Rationalization to Find Limits
    Math Deficiency – IITopic 14 of 32

    Substitution to Find Limits, Rationalization to Find Limits

    9 minread
    1,459words
    Intermediatelevel

    Substitution to Find Limits

    Substitution is a basic method for finding the limit of a function as xxx approaches a certain value. If a function is continuous at a point, you can directly substitute the value of xxx into the function to find the limit. This is the simplest and most straightforward approach when dealing with limits.

    Steps for Substitution:

    1. Check Continuity: If the function is continuous at the point you're evaluating, you can substitute the value of xxx directly into the function.
    2. Substitute x=cx = cx=c: Simply replace xxx with the value that xxx is approaching (denoted as ccc) in the function.

    Example 1: Direct Substitution

    Find the limit of the function f(x)=3x+2f(x) = 3x + 2f(x)=3x+2 as xxx approaches 4:

    lim⁡x→4(3x+2)\lim_{x \to 4} (3x + 2)x→4lim​(3x+2)

    Since the function is a polynomial (which is continuous everywhere), we can substitute x=4x = 4x=4 directly:

    f(4)=3(4)+2=12+2=14f(4) = 3(4) + 2 = 12 + 2 = 14f(4)=3(4)+2=12+2=14

    Thus,

    lim⁡x→4(3x+2)=14\lim_{x \to 4} (3x + 2) = 14x→4lim​(3x+2)=14

    Example 2: Direct Substitution (Rational Function)

    Now, consider a rational function:

    f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1​

    Find the limit as x→1x \to 1x→1:

    lim⁡x→1x2−1x−1\lim_{x \to 1} \frac{x^2 - 1}{x - 1}x→1lim​x−1x2−1​

    Directly substituting x=1x = 1x=1 into the function:

    f(1)=12−11−1=00f(1) = \frac{1^2 - 1}{1 - 1} = \frac{0}{0}f(1)=1−112−1​=00​

    This results in an indeterminate form 00\frac{0}{0}00​, so direct substitution doesn’t work. In such cases, we need to simplify the expression further, which leads us to the next method: Rationalization.


    Rationalization to Find Limits

    Rationalization is a method used to simplify expressions involving square roots or complex rational expressions, particularly when direct substitution results in an indeterminate form like 00\frac{0}{0}00​.

    The goal of rationalization is to manipulate the function algebraically to eliminate the square roots (or other complex terms) and make the limit easier to evaluate.

    Steps for Rationalization:

    1. Identify the Indeterminate Form: When direct substitution results in indeterminate forms like 00\frac{0}{0}00​, rationalizing can help.
    2. Multiply by the Conjugate: If the expression contains square roots, multiply both the numerator and the denominator by the conjugate of the expression.
    3. Simplify: After multiplying by the conjugate, simplify the expression to remove the indeterminate form and find the limit.

    Example 1: Rationalizing a Square Root Expression

    Consider the following limit:

    lim⁡x→4x−2x−4\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}x→4lim​x−4x​−2​
    1. Substitute x=4x = 4x=4 directly:
    4−24−4=2−20=00\frac{\sqrt{4} - 2}{4 - 4} = \frac{2 - 2}{0} = \frac{0}{0}4−44​−2​=02−2​=00​

    This results in an indeterminate form, so we must rationalize the numerator by multiplying by the conjugate of x−2\sqrt{x} - 2x​−2.

    1. Multiply the numerator and denominator by the conjugate:
    x−2x−4×x+2x+2\frac{\sqrt{x} - 2}{x - 4} \times \frac{\sqrt{x} + 2}{\sqrt{x} + 2}x−4x​−2​×x​+2x​+2​

    This gives:

    (x−2)(x+2)(x−4)(x+2)=x−4(x−4)(x+2)\frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x - 4)(\sqrt{x} + 2)} = \frac{x - 4}{(x - 4)(\sqrt{x} + 2)}(x−4)(x​+2)(x​−2)(x​+2)​=(x−4)(x​+2)x−4​
    1. Simplify:

    We can cancel out x−4x - 4x−4 from the numerator and denominator (note that this is valid for x≠4x \neq 4x=4):

    1x+2\frac{1}{\sqrt{x} + 2}x​+21​
    1. Substitute x=4x = 4x=4 again:

    Now, substitute x=4x = 4x=4 into the simplified expression:

    14+2=12+2=14\frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}4​+21​=2+21​=41​

    Thus, the limit is:

    lim⁡x→4x−2x−4=14\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} = \frac{1}{4}x→4lim​x−4x​−2​=41​

    Example 2: Rationalizing with a Difference of Squares

    Consider the limit:

    lim⁡x→3x2−9x−3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}x→3lim​x−3x2−9​
    1. Substitute x=3x = 3x=3 directly:
    32−93−3=9−90=00\frac{3^2 - 9}{3 - 3} = \frac{9 - 9}{0} = \frac{0}{0}3−332−9​=09−9​=00​

    Again, we get an indeterminate form, so we simplify the expression by factoring the numerator.

    1. Factor the numerator:
    x2−9=(x−3)(x+3)x^2 - 9 = (x - 3)(x + 3)x2−9=(x−3)(x+3)

    Thus, the expression becomes:

    (x−3)(x+3)x−3\frac{(x - 3)(x + 3)}{x - 3}x−3(x−3)(x+3)​
    1. Cancel out x−3x - 3x−3:

    We can cancel x−3x - 3x−3 from the numerator and denominator (for x≠3x \neq 3x=3):

    f(x)=x+3f(x) = x + 3f(x)=x+3
    1. Substitute x=3x = 3x=3:

    Now, substitute x=3x = 3x=3:

    f(3)=3+3=6f(3) = 3 + 3 = 6f(3)=3+3=6

    Thus, the limit is:

    lim⁡x→3x2−9x−3=6\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6x→3lim​x−3x2−9​=6

    Summary of Techniques

    • Substitution: If a function is continuous at a point, you can directly substitute the value of xxx into the function to find the limit.
    • Rationalization: If substitution leads to an indeterminate form (like 00\frac{0}{0}00​), rationalizing the numerator or denominator (especially when square roots are involved) can help simplify the expression, making it easier to evaluate the limit.
    Previous topic 13
    Limit: Notation, Graphs to Find Limits, Tables to Find Limits
    Next topic 15
    One Sided Limits and Continuity

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