Rank the following functions in order of how quickly they grow as x \rightarrow \infty a....


Rank the following functions in order of how quickly they grow as {eq}x \rightarrow \infty {/eq}

{eq}a. y_1 = x^4 \\ b. y_2 = e^{-x} \\ c. y_3 = \ln x \\ d. y_4 = \sqrt x \\ e. y_5 = 5^x {/eq}

Rate of Growth

To determine which function is faster as the independent variable gets larger, we will evaluate the limit of the ratio of the functions.

If {eq}\displaystyle \lim_{x\to \infty} \frac{f(x)}{g(x)}= \text{nonzero finite value} \implies \text{ they have the same rate}\\ \displaystyle \lim_{x\to \infty} \frac{f(x)}{g(x)}=0 \implies g(x) \text{ grows faster than } f(x)\\ \displaystyle \lim_{x\to \infty} \frac{f(x)}{g(x)}=\infty \implies f(x) \text{ grows faster than } g(x)\\ {/eq}

Tangent Lines to a Graph

The tangent line to the graph of a function {eq}\displaystyle f(x) {/eq} at the point {eq}\displaystyle a {/eq}

is given by {eq}\displaystyle y=f(a)+f'(a)(x-a), {/eq} where {eq}\displaystyle '=\frac{d}{dx}. {/eq}

Answer and Explanation:

To determine which function from {eq}\displaystyle y_1(x)=x^{4}, y_2(x)=e^{-x}, y_3(x)=\ln x, y_4(x)=\sqrt{x}, y_5(x)=5^x {/eq} grows the fastest as x goes to infinity, we will evaluate the following limits.

{eq}\displaystyle \begin{align} \lim_{x\to \infty} \frac{y_1(x)}{y_2(x)}&=\lim_{x\to \infty} \frac{x^{4}}{e^{-x}}\\ &=\lim_{x\to \infty} x^{4}e^{x}\\ &=\infty \\ &\text{ therefore the polynomial function }y_1(x)=x^{4}\text{ grows faster than the exponential } e^{-x} \text{ as } x\to \infty\\ \lim_{x\to \infty} \frac{y_1(x)}{y_3(x)}&=\lim_{x\to \infty} \frac{x^{4}}{\ln x}\\ &\overset{\frac{\infty}{\infty}}{=} \lim_{x\to \infty} \frac{\frac{d}{dx}\left(x^{4}\right)}{\frac{d}{dx}\left(\ln x\right)}&\left[\text{applying L'Hospital rule}\right]\\ &=\lim_{x\to \infty} \frac{4x^{3}}{\frac{1}{x}}\\ &=\lim_{x\to \infty} 4x^{4}\\ &=\infty\\ &\text{ therefore, the polynomial function } y_1(x)=x^{4} \text{ is faster than the logarithmic function } \ln x.\\ \lim_{x\to \infty} \frac{y_1(x)}{y_4(x)}&=\lim_{x\to \infty}\frac{x^{4}}{\sqrt{x}}\\ &=\lim_{x\to \infty}\frac{x^{4}}{x^{1/2}}\\ &=\lim_{x\to \infty} x^{7/2}\\ &=\infty\\ &\text{ therefore, the polynomial function } x^{4} \text{ is faster than the square root function } \sqrt{x}.\\ \lim_{x\to \infty} \frac{y_1(x)}{y_5(x)}&=\lim_{x\to \infty} \frac{x^{4}}{5^x}\\ &\overset{\frac{\infty}{\infty}}{=} \lim_{x\to \infty} \frac{\frac{d}{dx}\left(x^{4}\right)}{\frac{d}{dx}\left(5^x\right)}=\lim_{x\to \infty} \frac{4x^{3}}{5^x \ln 5}\\ &=\lim_{x\to \infty} \frac{4\cdot 3\cdot 2}{5^x (\ln 5)^4}, &\left[\text{applying L'Hospital rule three more times}\right]\\ &=\frac{4!}{(\ln 5)^4}\lim_{x\to \infty} \frac{1}{5^x}\\ &=0\\ &\text{ therefore, the exponential function } y_5(x)=5^{x} \text{ is faster than the polynomial function } x^4.\\\\ \implies &\boxed{\text{ the function }y_5(x)=5^{x} \text{ grows the fastest as }x\to \infty, \text{ option e)}}. \end{align} {/eq}

Learn more about this topic:

Applying L'Hopital's Rule in Simple Cases

from Math 104: Calculus

Chapter 9 / Lesson 10

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