Problem 7 - Calculus Early Transcendentals

Question

Compare the growth rates of \(\left\\{n^{100}\right\\}\) and \(\left\\{e^{n / 100}\right\\}\) as \(n \rightarrow \infty\).

Step-by-Step Solution

Answer

Answer: The function \(e^{\frac{n}{100}}\) grows faster than \(n^{100}\) as \(n\) approaches infinity.

Step 1: Write the Ratio

Write the ratio of the two functions: \(\frac{n^{100}}{e^{\frac{n}{100}}}\).

Step 2: Evaluate the limit

Evaluate the limit of the ratio as \(n\) approaches infinity: $$\lim_{n \to \infty} \frac{n^{100}}{e^{\frac{n}{100}}}$$

Step 3: Use L'Hôpital's Rule

Since we have an indeterminate form (\(\frac{\infty}{\infty}\)), we can apply L'Hôpital's Rule. Differentiate the numerator and denominator with respect to \(n\): $$\frac{d}{dn}\left(n^{100}\right) = 100 n^{99}$$ $$\frac{d}{dn}\left(e^{\frac{n}{100}}\right) = \frac{1}{100} e^{\frac{n}{100}}$$ Now, compute the limit of the ratio of the derivatives: $$\lim_{n \to \infty} \frac{100 n^{99}}{\frac{1}{100} e^{\frac{n}{100}}}$$

Step 4: Simplify the expression

Simplify the expression by multiplying both the numerator and the denominator by 100: $$\lim_{n \to \infty} \frac{10000 n^{99}}{e^{\frac{n}{100}}}$$

Step 5: Apply L'Hôpital's Rule repeatedly

Since we still have an indeterminate form (\(\frac{\infty}{\infty}\)), continue to apply L'Hôpital's Rule repeatedly until we obtain either a finite nonzero limit or a determinate form. After 99 more applications, we get: $$\lim_{n \to \infty} \frac{100^{100} \cdot 100!}{e^{\frac{n}{100}}}$$

Step 6: Evaluate the final limit

As \(n\) approaches infinity, the exponential term in the denominator \(e^{\frac{n}{100}}\) will dominate, and the limit becomes: $$\lim_{n \to \infty} \frac{100^{100} \cdot 100!}{e^{\frac{n}{100}}} = 0$$

Step 7: Interpret the result

Since the limit is 0, we can conclude that the growth rate of \(n^{100}\) is slower than the growth rate of \(e^{\frac{n}{100}}\) as \(n\) approaches infinity.