Question

Lim (2n^2+3)/n^2-4 n->бесконечность

Answer 1

$\lim_{n \to \infty} \frac{2n^2+3}{n^2-4} = \\ n^2=n^2 \\ \frac{2}{1} = 2$

Answer 2

Solving the Limit

To solve the limit lim((2n^2+3)/(n^2-4+n)) as n approaches infinity, we can use algebraic manipulation and the concept of limits.

First, let's rewrite the expression to simplify it: lim((2n^2+3)/(n^2-4+n))

We can simplify the expression by dividing every term by the highest power of n in the denominator. This will help us to determine the behavior of the expression as n approaches infinity.

Dividing every term by n^2, we get: lim((2 + 3/n^2)/(1 - 4/n + 1/n^2))

As n approaches infinity, 3/n^2 and 4/n will approach 0, and 1/n^2 will approach 0 faster than 4/n. Therefore, the expression simplifies to: lim((2 + 0)/(1 - 0 + 0)) lim(2/1) 2

So, the limit of the given expression as n approaches infinity is 2.