I am trying to prove the Theorem in the title. I need help verifying my solution.
For all real numbers x, we have x≤−5 if and only if 1≤2x+3x−2≤2.
Proof. To prove the implication, by adding x+3 to both sides of x≤−5, one obtains
2x+3≤x−2
Since x≤−5, one can divide both sides of the inequality and obtains,
2x+3x−2≥1
By taking the limit of 2x+3x−2 as x approaches −[infinity],
lim
Hence, 1 \leq \frac{2x+3}{x-2} \leq 2.
To prove the converse, observe that if x-2 \geq 0, by multiplying both sides of 1 \leq \frac{2x+3}{x-2} \leq 2 by x-2, one obtains
\begin{align}
x-2 \leq 2x+3 \leq 2x -4\\
-2 \leq x+3 \leq x-4\\
-5 \leq x \leq x-7
\end{align}
Since no non-negative x satisfies the above inequalities, x-2 can not be greater than or equal to 0.
In case that x-2 < 0 by multiplying both sides by x-2, one obtains
\begin{align}
x-2 \geq 2x+3 \geq 2x -4\\
-2 \geq x+3 \geq x-4\\
-5 \geq x \geq x-7
\end{align}
Because all x \in (-\infty , -5] satisfy the above inequalities, x-2 < 0 and x \leq -5.
Since the implication and its converse are both true, x\leq -5 if and only if 1 \leq \frac{2x+3}{x-2} \leq 2.
