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hanmoyuan

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Find the limit of [tex]\frac{2^n}{n!}[/tex] when n approaches infinity.

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In summary, the limit of 2^n/n! as n approaches infinity is equal to 0. This can be calculated using the ratio test, and the expression 2^n/n! represents the sequence of terms in the infinite series for the function f(x) = 2^x. The reason for this limit being 0 is because the denominator grows much faster than the numerator, and it has significance in understanding the behavior of exponential functions for large values of x.

- #1

hanmoyuan

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Find the limit of [tex]\frac{2^n}{n!}[/tex] when n approaches infinity.

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Mark44

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What have you tried? Per the rules of this forum (see the Rules button near the top of the page), you need to show an effort at solving your problem before we can help.hanmoyuan said:Find the limit of [tex]\frac{2^n}{n!}[/tex] when n approaches infinity.

The limit of 2^n/n! as n approaches infinity is equal to 0. This means that as n gets bigger and bigger, the value of 2^n/n! will get closer and closer to 0.

To calculate this limit, we can use the ratio test. This involves taking the limit of the ratio of the n+1 term to the nth term. In this case, the ratio would be (2^(n+1)/(n+1)!)/(2^n/n!), which simplifies to 2/(n+1). As n approaches infinity, this ratio will approach 0, indicating that the limit is also 0.

The expression 2^n/n! represents the sequence of terms that make up the infinite series for the function f(x) = 2^x. As n approaches infinity, the terms in this sequence get smaller and smaller, eventually approaching 0.

This limit equals 0 because the denominator (n!) grows much faster than the numerator (2^n) as n approaches infinity. This means that the value of the fraction will become smaller and smaller, approaching 0 as n gets larger.

The significance of this limit is that it helps us understand the behavior of exponential functions as the input (in this case, n) gets larger and larger. It also allows us to approximate the value of these functions for very large values of x.

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