Limits

Understand the intuitive concept of the limit of a function

  • Theory
  • Examples
    Find \(\mathop {\lim }\limits_{x \to 5} \;\frac{{\sqrt {{x^2} + 5x} - \sqrt {50} }}{{\sqrt {{x^2} - 10} - \sqrt {15} }}\).
  • Solutions
    \(\begin{array}{c}\mathop {\lim }\limits_{x \to 5} \;\frac{{\sqrt {{x^2} + 5x} - \sqrt {50} }}{{\sqrt {{x^2} - 10} - \sqrt {15} }}
    = \mathop {\lim }\limits_{x \to 5} \;\frac{{(\sqrt {{x^2} + 5x} - \sqrt {50} )(\sqrt {{x^2} + 5x} + \sqrt {50} )(\sqrt {{x^2} - 10} + \sqrt {15} )}}{{(\sqrt {{x^2} - 10} - \sqrt {15} )(\sqrt {{x^2} + 5x} + \sqrt {50} )(\sqrt {{x^2} - 10} + \sqrt {15} )}}\\
    = \mathop {\lim }\limits_{x \to 5} \;{\frac{{[({x^2} + 5x) - 50](\sqrt {{x^2} - 10} + \sqrt {15} )}}{{[({x^2} - 10) - 15](\sqrt {{x^2} + 5x} + \sqrt {50} )}}^{^{}}}\\
    = \mathop {\lim }\limits_{x \to 5} \;{\frac{{({x^2} + 5x - 50)(\sqrt {{x^2} - 10} + \sqrt {15} )}}{{({x^2} - 25)(\sqrt {{x^2} + 5x} + \sqrt {50} )}}^{^{}}}\\
    = \mathop {\lim }\limits_{x \to 5} \;{\frac{{(x + 10)(x - 5)(\sqrt {{x^2} - 10} + \sqrt {15} )}}{{(x + 5)(x - 5)(\sqrt {{x^2} + 5x} + \sqrt {50} )}}^{^{}}}\\
    = \mathop {\lim }\limits_{x \to 5} \;{\frac{{(x + 10)(\sqrt {{x^2} - 10} + \sqrt {15} )}}{{(x + 5)(\sqrt {{x^2} + 5x} + \sqrt {50} )}}^{^{}}}\\
    = {\frac{{(5 + 10)(\sqrt {{5^2} - 10} + \sqrt {15} )}}{{(5 + 5)[\sqrt {{5^2} + 5(5)} + \sqrt {50} ]}}^{^{}}}\\
    = {\frac{{30\sqrt {15} }}{{20\sqrt {50} }}^{^{}}}\\
    = \frac{{3\sqrt {15} }}{{2 \times 5\sqrt 2 }} \times {\frac{{\sqrt 2 }}{{\sqrt 2 }}^{^{}}}\\
    = {\underline{\underline {\frac{{3\sqrt {30} }}{{20}}}} ^{^{}}}\end{array}\)

Find the limit of a function

  • Theory
    [codemissing] lim θ
  • Examples
    Find \(\mathop {\lim }\limits_{x \to 0} \;\frac{{\sin (1 - {e^{2x}})}}{{1 - {e^x}}}\).
  • Solutions
    \(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \;\frac{{\sin (1 - {e^{2x}})}}{{1 - {e^x}}}
    = \mathop {\lim }\limits_{x \to 0} \;\frac{{\sin (1 - {e^{2x}})}}{{1 - {e^x}}} \cdot \frac{{1 + {e^x}}}{{1 + {e^x}}}\\
    = \mathop {\lim }\limits_{x \to 0} \;\frac{{(1 + {e^x})\sin (1 - {e^{2x}})}}{{1 - {e^{2x}}}}\end{array}\)

    Let \(t = 1 - {e^{2x}}\), then \(x = \frac{1}{2}\ln (1 - t)\).
    When \(x \to 0\), \(t \to 0\).

    \(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \;\frac{{\sin (1 - {e^{2x}})}}{{1 - {e^x}}}
    = \mathop {\lim }\limits_{t \to 0} \;\frac{{[1 + {e^{\frac{1}{2}\ln (1 - t)}}]\sin t}}{t}\\
    = \mathop {\lim }\limits_{t \to 0} \;[1 + {e^{\frac{1}{2}\ln (1 - t)}}] \cdot \mathop {\lim }\limits_{t \to 0} \;\frac{{\sin t}}{t}\\
    = [1 + {e^{\frac{1}{2}\ln (1 - 0)}}] \cdot 1\\
    = \underline{\underline 2} \end{array}\)