Question

A) ) lim n infty (2n+1)(n-3) (n+2)(n+4)

Answer 1

$\lim_{n \to \infty} \frac{(2n + 1)(n - 3)}{(n + 2)(n + 4)} = \lim_{n \to \infty} \frac{2 {n}^{2} - 6n + n - 3 }{ {n}^{2} + 4n + 2n + 8} = \lim_{n \to \infty} \frac{2 {n}^{2} - 5n - 3 }{ {n}^{2} + 6n + 8} = \lim_{n \to \infty} \frac{ {n}^{2} (2 - \frac{5}{n} - \frac{3}{ {n}^{2} } ) }{ {n}^{2}(1 + \frac{6}{n} + \frac{8}{ {n}^{2} }) } = \frac{2}{1} = 2$

Ответ: 2

Answer 2

To find the limit of the given expression as n approaches infinity, you can use the concept of limits and simplify the expression. Here's how you can do it:

lim (n → ∞) [(2n+1)(n-3) / (n+2)(n+4)]

First, let's consider the highest degree terms in the numerator and denominator, as those will dominate as n approaches infinity.

In the numerator, the highest degree term is 2n * n, which is 2n^2.

In the denominator, the highest degree term is n * n, which is n^2.

Now, take the ratio of these leading coefficients:

(2n^2) / (n^2) = 2

As n approaches infinity, the limit of a constant like 2 is still 2. So, the limit of the given expression as n approaches infinity is 2:

lim (n → ∞) [(2n+1)(n-3) / (n+2)(n+4)] = 2