if-lim-x-a-6-x-1-ax-1-2-find-the-largest-and-smallest-correct-value-for-a-

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 AST 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 ⩽ \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 AST 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 AST 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 AST 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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