First slide

Introduction to limits

Question

If $f (x) = \lim_{x \to \infty} \frac{1}{{(\frac{3}{π} \tan^{- 1} 2 x)}^{2 n} + 5}$ . Then the set of values of x for which f(x) = 0 is

Moderate

A

$|2 x| > \sqrt{3}$
B

$|2 x| < \sqrt{3}$
C

$|2 x| \geq \sqrt{3}$
D

$|2 x| \leq \sqrt{3}$

Solution

$\begin{array}{l} Given g (x) = \lim_{x \to \infty} \frac{1}{{(\frac{3}{π} \tan^{- 1} 2 x)}^{2 n} + 5} = 0 \\ or \lim_{x \to \infty} {[{(\frac{3}{π} \tan^{- 1} 2 x)}^{2}]}^{n} \to \infty \\ or {(\frac{3}{π} \tan^{- 1} 2 x)}^{2} > 1 \\ or |\tan^{- 1} 2 x| > \frac{π}{3} \\ i . e ., \tan^{- 1} 2 x < - \frac{π}{3} or \tan^{- 1} 2 x > \frac{π}{3} \\ i . e ., 2 x < - \sqrt{3} or 2 x > \sqrt{3}, i . e ., |2 x| > \sqrt{3} \end{array}$

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