Limit Theorem and Euler's Numbers

Limit Theorem

The properties of the limit of a function can be summarized in the Limit Theorem as follows.

1. If $f\left( x \right) = k$ then $\mathop {\lim }\limits_{x \to a} f\left( x \right) = k$, for every $k$ constant and $a$ real numbers.
2. If $f\left( x \right) = x$ then $\mathop {\lim }\limits_{x \to a} f\left( x \right) = a$, for every $a$ number real.
3. $\mathop {\lim }\limits_{x \to a} \left\{ {f\left( x \right) \pm g\left( x \right)} \right\} = \mathop {\lim }\limits_{x \to a} f\left( x \right) \pm \mathop {\lim }\limits_{x \to a} g\left( x \right)$
4. If $k$ is a constant then $\mathop {\lim }\limits_{x \to a} kf\left( x \right) = k\mathop {\lim }\limits_{x \to a} f \left( x \right)$
5. $\mathop {\lim }\limits_{x \to a} \left\{ {f\left( x \right).g\left( x \right)} \right\} = \mathop {\lim }\limits_{x \to a} f\left( x \right).\mathop {\lim }\limits_{x \to a} g\left( x \right)$
6. $\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\ mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}$ with $\mathop {\lim }\limits_{x \to a} g\left( x \right) \ne 0$.
7. $\mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^2} = {\left[ {\mathop {\lim }\limits_{x \to a} f\left( x \right)} \right]^2}$
8. $\mathop {\lim }\limits_{x \to a} \sqrt[n]{{f\left( x \right)}} = \sqrt[n]{{\mathop {\lim }\ limits_{x \to a} f\left( x \right)}}$ with $\mathop {\lim }\limits_{x \to a} f\left( x \right) \ge 0$ for $n$ even.

Example

Calculate the limit value of the following function.

1. $\mathop {\lim }\limits_{x \to 1} \left( {2{x^2} - 3x + 5} \right)$
2. $\mathop {\lim }\limits_{x \to 0} \sqrt {\frac{{{x^2} + 4}}{{x + 4}}} $

Answer

$\begin{array}{l} \mathop {\lim }\limits_{x \to 1} \left( {2{x^2} - 3x + 5} \right) &= \mathop {\lim }\limits_{x \to 1} 2{x^2} - \mathop {\lim }\limits_{x \to 1} 3x + \mathop {\lim }\limits_{x \to 1} 5\\ &= 2\mathop {\lim }\limits_{x \to 1} {x^2} - 2\mathop {\lim }\limits_{x \to 1} x + \mathop {\lim }\limits_{x \to 1} 5\\ &= 2{\left[ {\mathop {\lim }\limits_{x \to 1} x} \right]^2} - 2\mathop {\lim }\limits_{x \to 1} x + \mathop {\lim }\limits_{x \to 1} 5\\ &= {2.1^2} - 2.1 + 5\\ &= 2 - 2 + 5\\ &= 5 \end{array}$

So, $\mathop {\lim }\limits_{x \to 1} \left( {2{x^2} - 3x + 5} \right) = 5$.

$\begin{array}{l} \mathop {\lim }\limits_{x \to 0} \sqrt {\frac{{{x^2} + 4}}{{x + 4}}} &= \frac{{\mathop {\lim }\limits_{x \to 0} \sqrt {{x^2} + 4} }}{{\mathop {\lim \sqrt {x + 4} }\limits_{x \to 0} }}\\ &= \frac{{\sqrt {\mathop {\lim }\limits_{x \to 0} \left( {{x^2} + 4} \right)} }}{{\sqrt {\mathop {\lim }\limits_{x \to 0} \left( {x + 4} \right)} }}\\ &= \frac{{\sqrt {\mathop {\lim }\limits_{x \to 0} \mathop {\left( {{x^2}} \right) + \lim }\limits_{x \to 0} \left( 4 \right)} }}{{\sqrt {\mathop {\lim }\limits_{x \to 0} \left( x \right)\mathop {\lim }\limits_{x \to 0} \left( 4 \right)} }}\\ &= \frac{{\sqrt {{{\left[ {\mathop {\lim x}\limits_{x \to 0} } \right]}^2}\mathop { + \lim }\limits_{x \to 0} \left( 4 \right)} }}{{\sqrt {\mathop {\lim }\limits_{x \to 0} \left( x \right) + \mathop {\lim }\limits_{x \to 0} \left( 4 \right)} }}\\ &= \frac{{\sqrt {{0^2} + 4} }}{{\sqrt {0 + 4} }}\\ &= \frac{{\sqrt 4 }}{{\sqrt 4 }}\\ &= \frac{2}{2}\\ &= 1 \end{array}$

So, $\mathop {\lim }\limits_{x \to 0} \sqrt {\frac{{{x^2} + 4}}{{x + 4}}} = 1$.

Euler's Number

Eulerian number $e=2.718281828459...$ is an irrational number obtained from a limit form for the variable $x$ approaching infinity. The number $e$ is formulated as:

\[\begin{array}{l} \boxed{\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = e}\\ or\\ \boxed{\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{1}{x}} \right)^{ - x}} = e} \end{array}\]

Example

Calculate the limit value of the following function.

1. $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{3}{x}} \right)^x}$
2. $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{{x^2}}}} \right)^{{x^2 } - 4x}}$

Answer

$\begin{array}{l} \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{3}{x}} \right)^x} = \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{3}{x}} \right)^{\frac{x}{3}.3}}\\ = \mathop {\lim }\limits_{x \to \infty } {\left\{ {{{\left( {1 + \frac{3}{x}} \right)}^{\frac{x} {3}}}} \right\}^3}\\ = \mathop {\lim }\limits_{x \to \infty } {\left\{ {{{\left( {1 + \frac{1}{{\frac{x}{3}}}} \right )}^{\frac{x}{3}}}} \right\}^3}\\ = {e^3} \end{array}$

So, $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{3}{x}} \right)^x} = {e^3}$ .

$\begin{array}{l} \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{{x^2}}}} \right)^{{x^2} - 4x}} &= \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{{x^2}}}}}{{{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{4x}}}}\\ &= \frac{{\mathop {\lim }\limits_{x \to \infty } {{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{{x^2}}}}}{{\mathop {\lim }\limits_{x \to \infty } {{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{4x}}}}\\ &= \frac{{\mathop {\lim }\limits_{x \to \infty } {{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{{x^2}}}}}{{\mathop {\lim }\limits_{x \to \infty } {{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{{x^2}.\frac{4}{x}}}}}\\ &= \frac{{\mathop {\lim }\limits_{x \to \infty } {{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{{x^2}}}}}{{\mathop {\lim }\limits_{x \to \infty } {{\left\{ {{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{{x^2}}}} \right\}}^{\frac{4}{x}}}}}\\ &= \frac{e}{{{e^0}}}\\ &= \frac{e}{1} = e \end{array}$

So $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{{x^2}}}} \right)^{{x ^2} - 4x}} = e$.