Operations for functions: Exponential and logarithmic functions
Growth of an exponential function
The graphs of the exponential functions #2^x# and #2^{-x}# are as follows:
All graphs of exponential functions have a horizontal asymptote at \(y=0\) and grow in either the positive or negative direction:
Asymptotic behavior of exponential functions
- If #a\gt1#, then #\displaystyle\lim_{x\to-\infty} a^x = 0# and #\displaystyle\lim_{x\to\infty} a^x = \infty#.
- If #0\lt a\lt1#, then #\displaystyle\lim_{x\to-\infty} a^x = \infty# and #\displaystyle\lim_{x\to\infty} a^x = 0#.
The two statements are related: the graph of the function \(f(x)=\left ( \frac{1}{a} \right )^x=a^{-x}\) (remember the properties of exponents) is obtained by reflecting the graph of the function \(f(x)=a^x\) around the vertical axis \(x=0\).
In fact, for \(a\gt1\), the function \(a^x\) tends to infinity exceedingly fast when \(x\) tends to infinity:
Exponential versus polynomial growth
For \(a\gt1\) and \(n\in\mathbb{N}\) \[\lim_{x\to\infty}\frac{x^n}{a^x}=0\] and \[\lim_{x\to\infty}\frac{a^x}{x^n}=\infty\]
In other words, the term \(a^x\) becomes much bigger than the polynomial expression \(x^n\) (for any given natural number \(n\)), when \(x\) tends to infinity (the term \(a^x\) "grows faster" than the term \(x^n\)). Notice that this holds regardless how close #a# is to #1# and regardless how large #n# is. This is one of the reasons why exponential growth attracts so much attention.
Intuitively, this can be seen by concluding that #x^{3}\cdot 2^{9 x}# in the denominator is the fastest growing term: exponential functions grow much faster than the power functions and the exponent of the exponential function in the denominator is greater than the exponent of the exponential function in the numerator.
A more concrete approach is the following:
\[\begin{array}{rcl}\displaystyle\lim_{x\to\infty}\frac{x^{4}\cdot 2^{6 x}}{x^{3}\cdot 2^{9 x} }&=&\displaystyle\lim_{x\to\infty}\frac{x^{1}}{2^{3 x}}\\
&&\phantom{xx}\color{blue}{\text{division of numerator and denominator by }x^{3}\cdot2^{6 x}}\\
&=&\displaystyle\lim_{x\to\infty}\frac{x^{1}}{{8}^ x}\\
&&\phantom{xx}\color{blue}{2^{a\cdot b} = (2^a)^b }\\
&=&0\\
&&\phantom{xx}\color{blue}{\text{ exponential versus polynomial growth rule} }\\
\end{array}\]
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