Compute inf and sup of a set A

Question

Compute the infimum and supremum of $A=\{f(x)=\frac{2x+1}{x+2}: x>-2\}$.
I try to do these following passages: since $f$ is derivable I compute $f'(x)=\frac{3}{x+2}>0$ and so from Monotone Convergence Theorem I can conclude that: $\lim_{x\to-2^+}f(x)=-\infty=infA$ and $\lim_{x\to+\infty}f(x)=2=supA$. All right?

Answer 1

Your sketch seems more or less correct, depending on which of these theorems you actually have access to in the class and how the final proof is written.

That being said, you can also so this straight from the definitions of inf and sup.

Your supposition is probably correct:

Let $N \in \mathbb{N}$ be any number. Can you prove that there exists an $x \in (-2,\infty)$ such that $f(x) < N$? If so, you have proven that the inf is indeed $-\infty$.

Now, for the sup, we first need to prove that 2 is an upper bound. That shouldn't be too hard. Then, you need to show that for any $\varepsilon>0$, there exists an $x \in (-2,\infty)$ such that $\lvert 2-f(x) \rvert < \varepsilon$.

With these two, you don't need all the machinery that your answer gave.

Answer 2

You can use the fact that $$ \frac{2x+1}{x+2} = 2-\frac{3}{x+2}. $$ Using this you can prove that the function is decreasing for $x>-2$ instead of using the derivative. Also with this you can calculate the limits.