Question

find \\(\lim_{n \to \infty} \left( \frac{3n + 3}{n + 2} - \frac{2n + 2}{2n + 4} \
ight)^2\\)

Explanation:

Step1: Simplify each fraction

For $\frac{3n+3}{n+2}$: divide numerator/denominator by $n$: $\frac{3+\frac{3}{n}}{1+\frac{2}{n}}$.
For $\frac{2n+2}{2n+4}$: divide numerator/denominator by $n$: $\frac{2+\frac{2}{n}}{2+\frac{4}{n}}$.

Step2: Compute limits of fractions

As $n\to\infty$, $\frac{3}{n},\frac{2}{n},\frac{4}{n}\to0$.
$\lim_{n\to\infty}\frac{3+\frac{3}{n}}{1+\frac{2}{n}}=3$; $\lim_{n\to\infty}\frac{2+\frac{2}{n}}{2+\frac{4}{n}}=1$.

Step3: Calculate inside the parentheses

$3 - 1 = 2$.

Step4: Square the result

$2^2 = 4$.

Answer:

4