Determine whether the sequence converges or diverges. If it converges, find the limit. (If an answer does not exist, enter DNE.)
limit n approaches infinity of an = e^(−6/sqrt(n))
clearly e^(-6/√n) -> e^(-6/∞) = e^0 = 1
Well, let's see if this sequence converges or just likes to wander off into the abyss.
The sequence is defined as an = e^(−6/sqrt(n)). As n approaches infinity, we can see that the square root of n also approaches infinity. And as the denominator of the exponent gets larger and larger, the value inside the exponential function approaches zero.
Now, since e^0 is equal to 1, this means that the limit of this sequence as n approaches infinity is 1. So, the sequence converges to a limit of 1.
That's one steady sequence. It knows where it's going and it's not afraid to get there!
To determine whether the sequence converges or diverges, we need to find the limit of the sequence as n approaches infinity.
The sequence is given by an = e^(-6/sqrt(n)).
As n approaches infinity, the term 1/sqrt(n) approaches 0. Therefore, the exponent in the sequence expression approaches 0 as well.
Using the limit property of exponential functions, we know that e^0 = 1.
Thus, the limit as n approaches infinity of the sequence an = e^(-6/sqrt(n)) is 1.
Therefore, the sequence converges and its limit is 1.
To determine whether the sequence converges or diverges and find the limit, we need to evaluate the expression
lim (n→∞) e^(−6/√(n))
To do this, we'll apply a few steps:
Step 1: Simplify the expression
We can rewrite the expression as:
lim (n→∞) e^(-6/n^(1/2))
Step 2: Take the limit
As n approaches infinity, the denominator √(n) approaches infinity as well. So, we have:
lim (n→∞) e^(-6/∞)
Step 3: Evaluate the limit
e^(-6/∞) is an indeterminate form. To determine the limit, we can rewrite it as:
lim (n→∞) e^(-6/n)
Now, as n approaches infinity, the exponent -6/n approaches 0. Therefore, the limit becomes:
lim (n→∞) e^(0)
The value of e^0 is 1. So, the limit of the sequence as n approaches infinity is:
lim (n→∞) e^(-6/√(n)) = 1
Therefore, the sequence converges and its limit is 1.