Question

Use Theorem 6 to find the limit of the following sequences or state that they diverge. $$\left\\{\frac{e^{n / 10}}{2^{n}}\right\\}$$

Short answer

Answer: The limit of the sequence $$\left\\{\frac{e^{n / 10}}{2^{n}}\right\\}$$ as $$n$$ approaches infinity is 0.

Step-by-step solution

Find the limit of the numerator

We need to find the limit of the numerator sequence, which is $$\lim_{n \to \infty} e^{n/10}$$. This is a simple limit since the exponential function grows without bound as its exponent approaches infinity. Therefore, the limit of the numerator is: $$\lim_{n \to \infty} e^{n/10} = \infty$$

Find the limit of the denominator

Next, we need to find the limit of the denominator sequence, which is $$\lim_{n \to \infty} 2^n$$. This is again a simple limit since the exponential function with a base greater than 1 also grows without bound as its exponent approaches infinity. Therefore, the limit of the denominator is: $$\lim_{n \to \infty} 2^n = \infty$$

Apply Theorem 6 to find the limit of the sequence

Now that we have found the limits of both the numerator ($$\infty$$) and denominator ($$\infty$$), we can apply Theorem 6 to find the limit of the given sequence as follows: $$\lim_{n \to \infty} \frac{e^{n / 10}}{2^{n}} = \frac{\lim_{n \to \infty} e^{n / 10}}{\lim_{n \to \infty} 2^{n}} = \frac{\infty}{\infty}$$ Since both the numerator and denominator approach infinity, the limit of the ratio of these sequences is indeterminate in this form. However, we can use L'Hopital's rule to find the limit if we transform the expression into a continuous form. First, let $$f(x) = e^{x/10}$$ and $$g(x) = 2^x$$, and let's find the derivatives of these functions: - $$f'(x) = \frac{1}{10} e^{x/10}$$ - $$g'(x) = \ln(2) * 2^x$$ Now, according to L'Hopital's rule, we can find the limit of their ratio when both functions approach infinity by finding the limit of the ratio of their derivatives: $$\lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \lim_{x \to \infty} \frac{\frac{1}{10} e^{x/10}}{\ln(2) * 2^x}$$ Substitute $$n$$ for $$x$$: $$\lim_{n \to \infty} \frac{\frac{1}{10} e^{n/10}}{\ln(2) * 2^n}$$ Since both the numerator and the denominator still approach infinity, we can apply L'Hopital's rule once more: - $$f''(n) = \frac{1}{100} e^{n/10}$$ - $$g''(n) = (\ln(2))^2 * 2^n$$ Now, we find the limit of the ratio of their second derivatives: $$\lim_{n \to \infty} \frac{f''(n)}{g''(n)} = \lim_{n \to \infty} \frac{\frac{1}{100} e^{n/10}}{(\ln(2))^2 * 2^n} = 0$$ So, after applying L'Hopital's rule twice, we find that the limit of the given sequence is 0. Therefore, the answer is: $$\lim_{n \to \infty} \frac{e^{n / 10}}{2^{n}} = 0$$

Key concepts

L'Hopital's Rule

When trying to find the limit of a sequence, especially when you encounter an indeterminate form like \( \frac{\infty}{\infty} \), L'Hopital's rule can be incredibly useful. This rule states that if the limit of a fraction results in an indeterminate form, you can take the derivatives of the numerator and denominator until the limit no longer results in an indeterminate form.
Simply said, if you have \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), you can use:

• Find \( f'(x) \) and \( g'(x) \).
• Calculate \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \).
• Repeat if necessary, until you find a determinate form.

In our problem, both the numerator \( e^{n/10} \) and denominator \( 2^n \) approached infinity, leading us to use L'Hopital’s rule for resolution.

Infinity Form

The term 'infinity form' often comes into play in sequence limits. This happens when both the numerator and denominator of a sequence approach infinity as \( n \) goes to infinity. Such a situation is known as an indeterminate form because it doesn't directly suggest a clear limit. 
In mathematical terms, an indeterminate form like \( \frac{\infty}{\infty} \) needs careful analysis to find an exact limit. This is exactly why we deploy tools like L'Hopital's Rule, which transforms this form by considering the growth rates of the numerator and denominator. In our exercise, both functions were exponential, but one exponent (\( 2^n \)) grew faster than the other (\( e^{n/10} \)), leading to a final resolved limit of the sequence.

Theorem 6

Theorem 6 likely refers to a criterion for determining sequence behavior as \( n \) approaches infinity. This theorem assists in comparing growth rates of functions and often reveals when a sequence diverges or converges.
In our context, this theorem guided us in initial stages to recognize that both the numerator and denominator headed towards infinity. However, by transforming these into functions suitable for L'Hopital’s rule, we evaluated the second derivative comparison. The theorem provided a logical foundation, while L'Hopital's rule filled in the computational gaps to confirm convergence to zero as the limit of the sequence.