Question

If \(i = \sqrt{-1}\), then \[ \lim_{n \to \infty} \frac{(n + 2i)(3 + 7in)}{(2 - i)(6n^2 + 1)} \] is equal to:

• A. \(-\frac{7}{5}\)
• B. \(\frac{14}{5} - \frac{7}{5}i\)
• C. \(\frac{7}{5} - \frac{14}{5}i\)
• D. \(-\frac{7}{30} + \frac{7}{15}i\)

Answer 1

The Correct Option is D

Let’s compute the numerator and denominator separately as \(n \to \infty\): \[ (n + 2i)(3 + 7in) = 3n + 6i + 7in^2 + 14i^2n = 7in^2 - 14n + 3n + 6i = 7in^2 - 11n + 6i \] \[ \text{Denominator} = (2 - i)(6n^2 + 1) \Rightarrow \text{As } n \to \infty, \lim = \frac{7in^2}{6n^2(2 - i)} = \frac{7i}{6(2 - i)} \cdot \frac{2 + i}{2 + i} = \frac{7i(2 + i)}{6(4 + 1)} = \frac{7i(2 + i)}{30} \] \[ = \frac{7(2i + i^2)}{30} = \frac{7(2i - 1)}{30} = -\frac{7}{30} + \frac{14}{30}i = -\frac{7}{30} + \frac{7}{15}i \]