First slide

Introduction to limits

Question

If $\lim_{x \to \infty} \frac{n {. 3}^{n}}{n {(x - 2)}^{n} + n {. 3}^{n + 1} - 3^{n}} = \frac{1}{3}$ then the range of x is (where $n \in N$ )

Moderate

A

[2, 5)
B

(1, 5)
C

(-1, 5)
D

$(- \infty . \infty)$

Solution

$\begin{array}{l} \lim_{x \to \infty} \frac{n {. 3}^{n}}{n {(x - 2)}^{n} + n {. 3}^{n + 1} - 3^{n}} = \frac{1}{3} \\ or \lim_{x \to \infty} \frac{1}{\frac{{(x - 2)}^{n}}{3^{n}} + 3 - \frac{1}{n}} = \frac{1}{3} (Dividing N^{t} and D^{r} by n \times 3^{n}) \\ For \lim_{x \to \infty} to be equal to \frac{1}{3}, \\ \lim_{x \to \infty} (\frac{x - 2}{3}) n \to 0 \\ ∴ 2 \leq x < 5 \end{array}$

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