Evaluate the followings: \( \begin{array}{ll}\text { (i) } \lim _{n \rightarrow \infty} \frac{3 n^{2}+4 n+7}{2+3 n+4 n^{2}} & \text { (ii) } \lim _{x \rightarrow 2} \frac{x^{2}-3 x+2}{x-2}\end{array} \)

For the first limit: (i) \( \lim _{n \rightarrow \infty} \frac{3 n^{2}+4 n+7}{2+3 n+4 n^{2}} \) To evaluate this limit as \( n \) approaches infinity, we can identify the leading terms in both the numerator and the denominator. The leading term in the numerator is \( 3n^2 \) and in the denominator is \( 4n^2 \). So we have: \[ \lim _{n \rightarrow \infty} \frac{3n^2 + 4n + 7}{2 + 3n + 4n^2} = \lim _{n \rightarrow \infty} \frac{3 + \frac{4}{n} + \frac{7}{n^2}}{\frac{2}{n^2} + \frac{3}{n} + 4} \] As \( n \) approaches infinity, the terms \( \frac{4}{n} \), \( \frac{7}{n^2} \), \( \frac{2}{n^2} \), and \( \frac{3}{n} \) all go to 0, giving us: \[ \frac{3 + 0 + 0}{0 + 0 + 4} = \frac{3}{4} \] So, the limit is \( \frac{3}{4} \). For the second limit: (ii) \( \lim _{x \rightarrow 2} \frac{x^{2}-3x+2}{x-2} \) To evaluate this limit, we first notice that the numerator can be factored: \[ x^2 - 3x + 2 = (x-1)(x-2) \] Thus, we rewrite the limit: \[ \lim _{x \rightarrow 2} \frac{(x-1)(x-2)}{x-2} \] For \( x \neq 2 \), we can cancel the term \( (x-2) \): \[ = \lim _{x \rightarrow 2} (x - 1) = 2 - 1 = 1 \] So, the limit is \( 1 \). Final answers: (i) \( \frac{3}{4} \) (ii) \( 1 \)