if lim_(x→+∞) [(((a+6)x+1)/(ax+1))]=2→ find the largest and smallest correct value for a.


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Question Number 204360 by esmaeil last updated on 14/Feb/24

$i f$ $\lim_{x \to + \infty} [\frac{(a + 6) x + 1}{a x + 1}] = 2 \to$ $f i n d t h e l a r g e s t a n d s m a l l e s t c o r r e c t$ $v a l u e f o r a .$
	Answered by deleteduser1 last updated on 14/Feb/24

	$= \underset{x \to + \infty}{l i m} [\frac{a + 6 + \frac{1}{x}}{a + \frac{1}{x}}] = \frac{a + 6}{a} = 2 \Rightarrow a = 6$
	Commented by esmaeil last updated on 14/Feb/24

	$t h a n k y o u . a n d s m a l l e s t v a l u e ?$
	Commented by esmaeil last updated on 14/Feb/24
	$2≤((a+6)/a)<3→1≤(6/a)⟨2→ 3<a≤6→ { ((a_l =6)),((a_s =4 (a is correct))) :}$
	$2 ⩽ \frac{a + 6}{a} < 3 \to 1 ⩽ \frac{6}{a} ⟨ 2 \to$ $3 < a ⩽ 6 \to {\begin{cases} a_{l} = 6 \\ a_{s} = 4 (a i s c o r r e c t) \end{cases}$
	Commented by esmaeil last updated on 14/Feb/24

	$[\frac{10}{4}] = 2$
	Commented by deleteduser1 last updated on 15/Feb/24

	$I t h o u g h t [] w a s j u s t a b r a c k e t, i f [x] i s f l o o r o f$ $x, t h e n [\frac{a + 6}{a}] = 2 < \frac{a + 6}{a} ⩽ 3 \Rightarrow 3 ⩽ a < 6$ $\Rightarrow m i n (a) = 3 a n d u p p e r b o u n d o f a = 6$
	Answered by mr W last updated on 15/Feb/24

	$l e t t = \frac{1}{x}$ $x \to + \infty \Rightarrow t \to 0^{+}$ $\lim_{x \to + \infty} [\frac{(a + 6) x + 1}{a x + 1}]$ $= \lim_{t \to 0^{+}} [\frac{a + 6 + t}{a + t}] = 2$ $\Rightarrow 2 < \frac{a + 6}{a} ⩽ 3$ $\Rightarrow 3 ⩽ a < 6$
	Commented by deleteduser1 last updated on 14/Feb/24

	$I t s e e m s 6 {d o e s n}^{'} t w o r k .$ $a = 6 \Rightarrow \underset{x \to + \infty}{l i m} [\frac{12 x + 1}{6 x + 1}] = \underset{x \to + \infty}{l i m} [\frac{2 (6 x + 1) - 1}{6 x + 1}]$ $= \underset{x \to + \infty}{l i m} [2 - \frac{1}{6 x + 1}] = 1, s i n c e 1 < 2 - \frac{1}{6 x + 1} < 2 a s$ $x \to + \infty .$ $E v e n t h o u g h \underset{x \to + \infty}{l i m} 2 - \frac{1}{6 x + 1} = 2, i t a p p r o a c h e s 2$ $f r o m t h e n e g a t i v e a x i s (2^{-}), s o a l l v a l u e s a r e$ $l e s s t h a n 2 .$
	Commented by mr W last updated on 14/Feb/24

	$y o u a r e r i g h t . 3 ⩽ a < 6 .$
	Answered by deleteduser1 last updated on 14/Feb/24

	$\underset{x \to + \infty}{l i m} [\frac{a x + 1 + 6 x}{a x + 1}] = \underset{x \to + \infty}{l i m} [\frac{6 x}{a x + 1} + 1] = 2$ $I f 6 x < a x + 1, t h e n \frac{6 x}{a x + 1} < 1 \Rightarrow [\frac{6 x}{a x + 1} + 1] < 2$ $\Rightarrow 6 x ⩾ a x + 1 \Rightarrow a ⩽ \frac{6 x - 1}{x} = 6 - \frac{1}{x} < 6$ $a < 6 \Rightarrow [\frac{6 x}{a x + 1} + 1] = [\frac{a x + 1 + (6 - a) x - 1}{a x + 1} + 1]$ $= [2 + \frac{(6 - a) x - 1}{a x + 1}]$ $\underset{x \to + \infty}{l i m} \frac{(6 - a) x - 1}{a x + 1} = \frac{6 - a}{a} = \frac{6}{a} - 1$ $\underset{x \to + \infty}{l i m} [\frac{a x + 1 + 6 x}{a x + 1}] = \underset{x \to + \infty}{l i m} [2 + \frac{(6 - a) x - 1}{a x + 1}] = 2$ $\underset{x \to + \infty}{l i m} \frac{(6 - a) x - 1}{a x + 1} ⩽ 1 \Rightarrow \frac{6}{a} - 1 ⩽ 1 \Rightarrow a ⩾ 3$ $\Rightarrow 3 ⩽ a < 6$
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