# How you solve this ?#lim_(n->oo)(5^(n)n!)/(2^(n)n^n)#

Using Stirling assymptotic approximation

then

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To solve the limit lim_(n->oo)(5^(n)n!)/(2^(n)n^n), we can use the ratio test.

First, let's simplify the expression by canceling out common factors between the numerator and denominator.

We can rewrite the expression as (5/2)^n * (n!)/(n^n).

Now, let's focus on the ratio of consecutive terms.

Taking the ratio of the (n+1)-th term to the n-th term, we get [(5/2)^(n+1) * ((n+1)!)/((n+1)^(n+1))] / [(5/2)^n * (n!)/(n^n)].

Simplifying this further, we have [(5/2)^(n+1) * (n+1)! * n^n] / [(5/2)^n * n! * (n+1)^(n+1)].

Now, we can cancel out common factors.

This simplifies to [(5/2) * n^n] / [(n+1)^(n+1)].

As n approaches infinity, the (n+1)-th term divided by the n-th term approaches 1/2.

Therefore, the limit of the given expression is 1/2.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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- How do you find the limit of #ln(lnt)# as #t->oo#?
- What is the limit of #(x^2)(e^x)# as x goes to negative infinity?

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