Without-using-calculus-find-stepwise-solution-lim-x-a-x-1-x-where-a-gt-1-

Question Number 1466 by Rasheed Soomro last updated on 10/Aug/15

W i t h o u t u s i n g c a l c u l u s f i n d s t e p w i s e s o l u t i o n :

\underset{x \to - \propto}{l i m} \frac{a^{x} - 1}{x} w h e r e a > 1

Answered by 123456 last updated on 11/Aug/15

a > 1, x < 0 \Rightarrow a^{x} < a^{0} = 1

a^{x} - 1 < 1 - 1 = 0

a^{x} - 1 < 0

\frac{a^{x} - 1}{x} > 0

0 < a^{x}

- 1 < a^{x} - 1

- \frac{1}{x} > \frac{a^{x} - 1}{x}

\lim_{x \to - \infty} \frac{1}{x} = 0

them by squeze theorem since

\lim_{x \to - \infty} - \frac{1}{x} = \lim_{x \to - \infty} 0 = 0

0 < \frac{a^{x} - 1}{x} < - \frac{1}{x}

then

\lim_{x \to - \infty} \frac{a^{x} - 1}{x} = 0

Commented by Rasheed Ahmad last updated on 11/Aug/15

1 s t l i n e : a^{x} < 1

\Rightarrow a^{x} - 1 < 1 - 1

\Rightarrow \frac{a^{x} - 1}{x} > \frac{1 - 1}{x} [∵ x < 0 i n e q u a l i t y

r e v e r s e d .]

But

2 n d l i n e : \frac{a^{x} - 1}{x} < \frac{1 - 1}{x} = 0 ? ? ?

[i n e q . n o t r e v e r s e d ?]

Pl explain I want to learn from you .

Commented by Rasheed Soomro last updated on 11/Aug/15

3 rd, 4 th and 5 th lines :

\frac{a^{x} - 1}{x} < 0

0 < a^{x}

\overset{?}{\Rightarrow} - \frac{1}{x} < \frac{a^{x} - 1}{x} < 0

As I think, may be wrong :

\frac{a^{x} - 1}{x} < 0 \Rightarrow \frac{a^{x}}{x} - \frac{1}{x} < 0

a^{x} > 0, x < 0 \Rightarrow \frac{a^{x}}{x} < 0 \overset{?}{\Rightarrow} - \frac{1}{x} < \frac{a^{x} - 1}{x} < 0

H o w - \frac{1}{x} < 0 ? W h i l e a c c o r d i n g t o y o u x < 0 ?

Commented by 123456 last updated on 11/Aug/15

sorry by the mistakes : v

i forgot these details, thank you :)

Commented by Rasheed Soomro last updated on 11/Aug/15

P l e a s e r e w r i t e y o u r a n s w e r : a m i s t a k e - f r e e a n s w e r .

A c t u a l l y I {d i d n}^{'} t m e a n t o c o m p l a i n a b o u t l a c k o f d e t a i l o f

s t e p s . I o n l y w a n t e d t o b e s a t i s f i e d a b o u t t h e l o g i c o f y o u r

s o l u t i o n .

I a m d e e p l y i n t e r e s t e d i n y o u r s o l u t i o n . I w i l l b e Thankful i f

y o u r e w r i t e y o u r a n s w e r f o r m e .

Commented by Rasheed Soomro last updated on 12/Aug/15

THANKS a lot! Now I think the logic is perfect .

Answered by 123456 last updated on 12/Aug/15

we have that for a > 1, a^{x} is crescent, then

x < 0 \Rightarrow a^{x} < a^{0} = 1 \Rightarrow a^{x} < 1

and then

a^{x} - 1 < 1 - 1 = 0 \Rightarrow a^{x} - 1 < 0 (- 1 in both side)

\frac{a^{x} - 1}{x} > 0 (\div x, and sonce x < 0, < turn into >)

we have that a^{x} > 0, then

a^{x} - 1 > - 1 (- 1 in both sides)

\frac{a^{x} - 1}{x} < - \frac{1}{x} (\div x, x < 0)

then

0 < \frac{a^{x} - 1}{x} < - \frac{1}{x}

then the squeze theorem work by this way

since

\frac{a^{x} - 1}{x} is bounced by 0 and - \frac{1}{x} and

0 \to 0 and - \frac{1}{x} \to 0 as x \to - \infty

then \frac{a^{x} - 1}{x} \to 0 as x \to - \infty

in formal way remember that

\lim_{x \to - \infty} f (x) = L

only if

\forall ϵ > 0, \exists M \in R, \forall x < M, ∣ f (x) - L ∣< ϵ

them

\lim_{x \to - \infty} 0 \Leftrightarrow M \in (- \infty, 0), x < M \Rightarrow∣ 0 - 0 ∣< ϵ

\lim_{x \to - \infty} - \frac{1}{x} = 0 \Leftrightarrow M ⩽ - \frac{1}{ϵ}, x < M

x < M < 0 \Rightarrow - ϵ ⩽ \frac{1}{M} < \frac{1}{x} < 0 \Rightarrow 0 < - \frac{1}{x} < - \frac{1}{M} ⩽ ϵ

∣ - \frac{1}{x} - 0 ∣< ϵ

and finaly, choose the same M above

\lim_{x \to - \infty} \frac{a^{x} - 1}{x} = 0

x < M

0 < \frac{a^{x} - 1}{x} < - \frac{1}{x} < - \frac{1}{M} ⩽ ϵ

∣ \frac{a^{x} - 1}{x} - 0 ∣< ϵ

as desired :)

Commented by Rasheed Ahmad last updated on 17/Aug/15

for a > 1, a^{x} is crescent \dots . .

what is the meaning of^{'} {crescent}^{'}

If a < 1 what term will be used ?

Commented by 123456 last updated on 28/Aug/15

a function is crescent if \forall x, y \in D (f)

x > y \Rightarrow f (x) > f (y)

for 0 < a < 1, a^{x} is decrescent, or in other

words

x < y \Rightarrow a^{x} > a^{y}

if a = 0, a^{x} is not well defined to x < 0

if a < 0, a^{x} is complex and multi valuated at x \in R / Z

Commented by Rasheed Ahmad last updated on 28/Aug/15

Thanks!

Answered by Rasheed Soomro last updated on 16/Aug/15

\frac{a^{x} - 1}{x} = (a^{x} - 1) \times \frac{1}{x}

\underset{x \to - \propto}{l i m} \frac{a^{x} - 1}{x} = \underset{x \to - \propto}{l i m} (a^{x} - 1) \times \underset{x \to - \propto}{l i m} (\frac{1}{x})

= (\underset{x \to - \propto}{l i m} a^{x} - \underset{x \to - \propto}{l i m} (1)) \times \underset{x \to - \propto}{l i m} (\frac{1}{x})

N o w {\underset{x \to - \propto}{l i m a}}^{x} = 0 f o r a > 1, \underset{x \to - \propto}{l i m} (1) = 1 a n d \underset{x \to - \propto}{l i m} (\frac{1}{x}) = 0

H e n c e

\underset{x \to - \propto}{l i m} \frac{a^{x} - 1}{x} = (0 - 1) \times 0 = 0

\underset{x \to - \propto}{l i m} \frac{a^{x} - 1}{x} = 0

Commented by 123456 last updated on 16/Aug/15

alright . :)

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