Finding limits algebraically - when direct substitution is not possible

Finding limits algebraically - when direct substitution is not possible

There are times when applying direct substitution would only give us an undefined solution. In this section, we will explore some cool tricks to evaluate limits algebraically, such as using conjugates, trigonometry, common denominators, and factoring.

Lessons

  • 1.
    Simplify Out "Zero Denominator" by Cancelling Common Factors

    Find limx3x29x3\lim_{x \to 3} \;\frac{{{x^2} - 9}}{{x - 3}}


  • 2.
    Expand First, Then Simplify Out "Zero Denominator" by Cancelling Common Factors

    Evaluate limh0(5+h)225h\lim_{h \to 0} \;\frac{{{{\left( {5 + h} \right)}^2} - 25}}{h}


  • 3.
    Simplify Out "Zero Denominator" by Rationalizing Radicals

    Evaluate:

    a)
    limx44x2x\lim_{x \to 4} \;\frac{{4 - x}}{{2 - \sqrt x }}
    (hint: rationalize the denominator by multiplying its conjugate)

    b)
    limx7x+23x7\lim_{x \to 7} \;\frac{{\sqrt {x + 2} - 3}}{{x - 7}}
    (hint: rationalize the numerator by multiplying its conjugate)


  • 4.
    Find Limits of Functions involving Absolute Value

    Evaluate limx0xx\lim_{x \to 0} \;\frac{{\left| x \right|}}{x}

    (hint: express the absolute value function as a piece-wise function)


  • 5.
    Find Limits Using the Trigonometric Identity:limθ0sinθθ=1\lim_{\theta \to 0} \;\frac{{{sin\;}\theta}}{{\theta}}=1

    Find limx0sin5x2x\lim_{x \to 0} \;\frac{{{sin\;}5x}}{{2x}}