Question

Find the limit of each of the following sequences, using L'Hôpital's rule when appropriate. $$ \frac{n^{2}}{2^{n}} $$

Short answer

The limit of the sequence is 0 as \( n \to \infty \).

Step-by-step solution

Identify the Sequence and its Components

The sequence given is \( a_n = \frac{n^2}{2^n} \). The numerator \( n^2 \) grows polynomially while the denominator \( 2^n \) grows exponentially.

Recognize the Form of the Limit Expression

The sequence approaches \( \infty/\infty \) as \( n \to \infty \), making it suitable for L'Hôpital's Rule, which can be applied to limits in the indeterminate form \( \frac{\infty}{\infty} \).

Convert the Sequence to a Limit Expression

Consider the limit as \( n \to \infty \): \[ \lim_{{n \to \infty}} \frac{n^2}{2^n} \]

Apply L'Hôpital’s Rule Once

Apply L'Hôpital's Rule by differentiating the numerator and denominator with respect to \( n \). The derivative of \( n^2 \) is \( 2n \), and the derivative of \( 2^n \) with respect to \( n \) is \( 2^n \log(2) \). Therefore, the expression becomes:\[ \lim_{{n \to \infty}} \frac{2n}{2^n \log(2)} \]

Apply L'Hôpital’s Rule Again

Apply L'Hôpital's Rule again as the expression \( \frac{2n}{2^n \log(2)} \) is still in the \( \frac{\infty}{\infty} \) form:The derivative of \( 2n \) is \( 2 \), and the derivative of \( 2^n \log(2) \) is \( 2^n \log^2(2) \). Therefore, the limit is:\[ \lim_{{n \to \infty}} \frac{2}{2^n \log^2(2)} \]

Evaluate the Limit

Now, evaluate the limit \( \lim_{{n \to \infty}} \frac{2}{2^n \log^2(2)} \). As \( n \to \infty \), \( 2^n \) grows exponentially, meaning the denominator grows much faster than the constant numerator 2. Therefore, the limit tends towards zero:\[ \lim_{{n \to \infty}} \frac{2}{2^n \log^2(2)} = 0 \]

Conclusion

The limit of the sequence \( \frac{n^2}{2^n} \) as \( n \to \infty \) is 0.

Key concepts

sequence limits

A sequence's limit describes what value the sequence's terms get closer to as they progress towards infinity. In the case of the given sequence, \( \frac{n^2}{2^n} \), it's crucial to recognize how the terms behave as \( n \) grows larger.

For this sequence:

• The numerator \( n^2 \) increases at a polynomial rate, meaning it's not very rapid compared to exponential growth.
• The denominator \( 2^n \) increases exponentially, which is incredibly fast, especially as \( n \) becomes large.

The speed of this growth tells us that \( 2^n \) will dominate \( n^2 \), pulling the entire fraction towards zero as \( n \to \infty \). Sequence limits help determine such end behaviors, revealing how the terms behave at infinity.

indeterminate forms

Indeterminate forms like \( \frac{\infty}{\infty} \) arise when evaluating limits, particularly those that don't immediately simplify to a clear value. Here, both the numerator and denominator become infinitely large as \( n \to \infty \).

When faced with \( \frac{n^2}{2^n} \):

• The expression fits the \( \frac{\infty}{\infty} \) form, meaning basic substitution won't work to find the limit.
• This situation of infinite quantities would, without tools, leave us uncertain of the sequence's limit.

L'Hôpital’s Rule is commonly used in calculus to address such expressions, converting them into more manageable limits with derivatives.

differentiation techniques

Differentiation is a technique in calculus that helps determine the rate at which a function changes. L'Hôpital's Rule employs this to handle limits of indeterminate forms like \( \frac{\infty}{\infty} \).

Applying L'Hôpital's Rule involves:

• Taking the derivative of the numerator \( n^2 \), which becomes \( 2n \).
• Taking the derivative of the denominator \( 2^n \), calculated as \( 2^n \log(2) \).

If the form \( \frac{\infty}{\infty} \) persists, as it did here, apply the rule again:

• The new numerator’s derivative \( 2n \) reduces to 2.
• The denominator’s derivative \( 2^n \log(2) \) turns into \( 2^n \log^2(2) \).

With a growing exponential denominator outpacing the constant numerator, differentiation techniques show the limit approaches zero.