Evaluate $\displaystyle \lim_{n \rightarrow \infty} \frac{n^2 - 5n}{\sqrt{3n^4 + 4n + 2}}$ or state that it diverges.
This is a sequence defined by a function, so how can we compute the limit?
This limit is equivalent to $\displaystyle \lim_{x \rightarrow \infty} \frac{x^2 - 5x}{\sqrt{3x^4 + 4x + 2}}$.
We did these limits in Calc 1. This is like the highest power rule, but you have to watch out for the square root.
$\frac{1}{\sqrt{3}}$
Evaluate $\displaystyle \lim_{n \rightarrow \infty} ne^{1/n}$ or state that it diverges.
This again is a sequence defined by a function, so how can we compute the limit?
This limit is equivalent to $\displaystyle \lim_{x \rightarrow \infty} xe^{1/x}$.
This limit is of the form $\infty \cdot 1$, so what does that mean for evaluating it?
This sequence diverges.
Write out the first 5 terms of the sequence $a_n = \frac{n^2}{n!}$, starting with $n=1$.
To write out each term of the sequence, we need to plug in consecutive values of $n$.
For $n=1$, we get $a_1 = \frac{1^2}{1!} = \frac{1}{1} = 1.$
$a_1 = 1$, $a_2 = 2$, $a_3 = \frac{3}{2}$. $a_4 = \frac{2}{3}$, $a_5 = \frac{5}{24}$.
Evaluate $\displaystyle \lim_{n \rightarrow \infty} \frac{3n^2 + 5n - \sin{n}}{4n^2 +1}$ or state that it diverges.
This is a sequence defined by a function, so we could use our methods from before. However, that $\sin{n}$ is problematic. Why?
Since $\lim_{n\rightarrow \infty} \sin{n}$ does not exist, we need to be more careful with it. The rest of the sequence is fine.
If we split off the $\frac{\sin{n}}{4n^2 + 1}$ part, the rest converges to $\frac{3}{4}$ with out an issue. What happens to this term? What theorem can we use to show that this limit exists?
If we use the Squeeze Theorem with $\pm \frac{1}{4n^2 + 1}$ as our bounding sequences, we get that this last part goes to zero.
$\frac{3}{4}$
Evaluate $\displaystyle \lim_{n \rightarrow \infty} \cos\left(\frac{n + \ln{n}}{n^2 - 2}\right)$ or state that it diverges.
This is a sequence inside a function, so you can separate out the function part first and just evaluate the limit of the sequence inside.
What happens to $\displaystyle \lim_{n\rightarrow \infty} \frac{n + \ln{n}}{n^2 - 2}$? You can split this into two terms to make it easier.
One of these terms needs a highest power rule, the other needs LHopital.
You should get that this limit is zero, which you then need to put inside the cosine function.
$1$
Write out the first $6$ terms of the sequence $b_n = n + \frac{1}{n}$.
Compute $\displaystyle \lim_{n \rightarrow \infty} \sqrt{n+3} - \sqrt{n}$. Hint: Multiply and divide by the conjugate and simplify.