The sequence $u_n$ given in the image is defined by:

$u_n = \frac{3}{2\sqrt{n} + 17}$

To find the limit of this sequence as $n \to \infty$, let's analyze the expression:

$u_n = \frac{3}{2\sqrt{n} + 17}$

As $n \to \infty$, the term $2\sqrt{n}$ grows much faster than the constant 17. Therefore, the expression $2\sqrt{n} + 17$ is dominated by $2\sqrt{n}$ for large values of $n$.

To find the limit, consider the asymptotic behavior:

$\lim_{n \to \infty} u_n = \lim_{n \to \infty} \frac{3}{2\sqrt{n} + 17}$

Since $2\sqrt{n}$ becomes very large as $n$ increases, the fraction approaches zero. Hence,

$\lim_{n \to \infty} u_n = 0$

Thus, the limit of the sequence $u_n$ as $n \to \infty$ is 0.

Would you like more details on the steps taken, or do you have any specific questions?

Related Questions:

1. How would the limit change if the sequence were $u_n = \frac{3}{2\sqrt{n} - 17}$?
2. What is the rate of convergence of the sequence $u_n = \frac{3}{2\sqrt{n} + 17}$ to its limit?
3. How do we compute the limit of a sequence where the term in the denominator grows exponentially, such as $u_n = \frac{3}{2^n + 17}$?
4. How does the presence of a square root affect the convergence of a sequence?
5. Can you find an $n$ after which $u_n$ becomes less than a given small positive number, say $0.01$?

Tip:

When finding limits of sequences, it's often useful to compare terms in the denominator to determine which dominates as $n$ approaches infinity.