### Limits

#### Understand the intuitive concept of the limit of a function

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

#### Find the limit of a function

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

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

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