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

Without using calculus find stepwise solution: lim_(x→−∝) ((a^x −1)/x) where a>1

$${Without}\:{using}\:{calculus}\:{find}\:{stepwise}\:{solution}: \\ $$$$\underset{{x}\rightarrow−\propto} {{lim}}\:\frac{{a}^{{x}} −\mathrm{1}}{{x}}\:{where}\:{a}>\mathrm{1} \\ $$

Answered by 123456 last updated on 11/Aug/15

a>1,x<0⇒a^x <a^0 =1 a^x −1<1−1=0 a^x −1<0 ((a^x −1)/x)>0 0<a^x −1<a^x −1 −(1/x)>((a^x −1)/x) lim_(x→−∞) (1/x)=0 them by squeze theorem since lim_(x→−∞) − (1/x)=lim_(x→−∞) 0=0 0<((a^x −1)/x)<−(1/x) then lim_(x→−∞) ((a^x −1)/x)=0

$${a}>\mathrm{1},{x}<\mathrm{0}\Rightarrow{a}^{{x}} <{a}^{\mathrm{0}} =\mathrm{1} \\ $$$${a}^{{x}} −\mathrm{1}<\mathrm{1}−\mathrm{1}=\mathrm{0} \\ $$$${a}^{{x}} −\mathrm{1}<\mathrm{0} \\ $$$$\frac{{a}^{{x}} −\mathrm{1}}{{x}}>\mathrm{0} \\ $$$$\mathrm{0}<{a}^{{x}} \\ $$$$−\mathrm{1}<{a}^{{x}} −\mathrm{1} \\ $$$$−\frac{\mathrm{1}}{{x}}>\frac{{a}^{{x}} −\mathrm{1}}{{x}} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}}=\mathrm{0} \\ $$$$\mathrm{them}\:\mathrm{by}\:\mathrm{squeze}\:\mathrm{theorem}\:\mathrm{since} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}−\:\frac{\mathrm{1}}{{x}}=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\mathrm{0}=\mathrm{0} \\ $$$$\mathrm{0}<\frac{{a}^{{x}} −\mathrm{1}}{{x}}<−\frac{\mathrm{1}}{{x}} \\ $$$$\mathrm{then} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{{a}^{{x}} −\mathrm{1}}{{x}}=\mathrm{0} \\ $$

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

1st line: a^x <1 ⇒ a^x −1<1−1 ⇒((a^x −1)/x) > ((1−1)/x) [ ∵ x<0 inequality reversed.] But 2nd line: ((a^x −1)/x)<((1−1)/x)=0 ??? [ ineq. not reversed?] Pl explain I want to learn from you.

$$\mathrm{1}{st}\:{line}:\:{a}^{{x}} <\mathrm{1} \\ $$$$\Rightarrow\:{a}^{{x}} −\mathrm{1}<\mathrm{1}−\mathrm{1} \\ $$$$\Rightarrow\frac{{a}^{{x}} −\mathrm{1}}{{x}}\:>\:\frac{\mathrm{1}−\mathrm{1}}{{x}}\:\:\:\:\:\:\left[\:\because\:{x}<\mathrm{0}\:\:{inequality}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{reversed}.\right] \\ $$$$\boldsymbol{\mathrm{But}} \\ $$$$\mathrm{2}{nd}\:{line}:\:\frac{{a}^{{x}} −\mathrm{1}}{{x}}<\frac{\mathrm{1}−\mathrm{1}}{{x}}=\mathrm{0}\:??? \\ $$$$\left[\:{ineq}.\:{not}\:{reversed}?\right] \\ $$$$\boldsymbol{\mathrm{Pl}}\:\boldsymbol{\mathrm{explain}}\:\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{want}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{learn}}\:\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{you}}. \\ $$

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

3rd,4th and 5th lines: ((a^x −1)/x)<0 0<a^x ⇒^? −(1/x)<((a^x −1)/x)<0 As I think,may be wrong : ((a^x −1)/x)<0 ⇒(a^x /x)−(1/x)<0 a^x >0 , x<0 ⇒ (a^x /x)<0 ⇒^? −(1/x)<((a^x −1)/x)<0 How −(1/x)<0 ? While according to you x<0?

$$\mathrm{3}\boldsymbol{\mathrm{rd}},\mathrm{4}\boldsymbol{\mathrm{th}}\:\boldsymbol{\mathrm{and}}\:\mathrm{5}\boldsymbol{\mathrm{th}}\:\boldsymbol{\mathrm{lines}}: \\ $$$$\frac{{a}^{{x}} −\mathrm{1}}{{x}}<\mathrm{0} \\ $$$$\mathrm{0}<{a}^{{x}} \\ $$$$\overset{?} {\Rightarrow}−\frac{\mathrm{1}}{{x}}<\frac{{a}^{{x}} −\mathrm{1}}{{x}}<\mathrm{0} \\ $$$$\boldsymbol{\mathrm{As}}\:\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{think}},\boldsymbol{\mathrm{may}}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{wrong}}\:: \\ $$$$\frac{{a}^{{x}} −\mathrm{1}}{{x}}<\mathrm{0}\:\Rightarrow\frac{{a}^{{x}} }{{x}}−\frac{\mathrm{1}}{{x}}<\mathrm{0} \\ $$$${a}^{{x}} >\mathrm{0}\:,\:{x}<\mathrm{0}\:\Rightarrow\:\:\frac{{a}^{{x}} }{{x}}<\mathrm{0}\:\overset{?} {\Rightarrow}−\frac{\mathrm{1}}{{x}}<\frac{{a}^{{x}} −\mathrm{1}}{{x}}<\mathrm{0}\: \\ $$$$\boldsymbol{\mathrm{H}}{ow}\:−\frac{\mathrm{1}}{{x}}<\mathrm{0}\:?\:\:{While}\:\:{according}\:{to}\:{you}\:{x}<\mathrm{0}? \\ $$

Commented by 123456 last updated on 11/Aug/15

sorry by the mistakes :v i forgot these details, thank you :)

$$\mathrm{sorry}\:\mathrm{by}\:\mathrm{the}\:\mathrm{mistakes}\::\mathrm{v} \\ $$$$\left.\mathrm{i}\:\mathrm{forgot}\:\mathrm{these}\:\mathrm{details},\:\mathrm{thank}\:\mathrm{you}\::\right) \\ $$

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

$${Please}\:{rewrite}\:\:\:{your}\:{answer}\::\:{a}\:{mistake}−{free}\:\:{answer}.\: \\ $$$${Actually}\:{I}\:{didn}'{t}\:{mean}\:{to}\:{complain}\:{about}\:{lack}\:{of}\:{detail}\:{of} \\ $$$${steps}.{I}\:{only}\:{wanted}\:{to}\:{be}\:{satisfied}\:{about}\:{the}\:{logic}\:{of}\:{your} \\ $$$${solution}. \\ $$$$\:{I}\:{am}\:{deeply}\:\:{interested}\:{in}\:{your}\:{solution}.{I}\:{will}\:{be}\:\boldsymbol{\mathrm{Thankful}}\:{if} \\ $$$${you}\:{rewrite}\:{your}\:{answer}\:{for}\:{me}. \\ $$

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

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

$$\boldsymbol{\mathrm{THANKS}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{lot}}!\:\boldsymbol{\mathrm{Now}}\:\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{think}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{logic}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{perfect}}. \\ $$

Answered by 123456 last updated on 12/Aug/15

we have that for a>1, a^x is crescent, then x<0⇒a^x <a^0 =1⇒a^x <1 and then a^x −1<1−1=0⇒a^x −1<0 (−1 in both side) ((a^x −1)/x)>0 (÷x, and sonce x<0, < turn into >) we have that a^x >0, then a^x −1>−1 (−1 in both sides) ((a^x −1)/x)<−(1/x) (÷x,x<0) then 0<((a^x −1)/x)<−(1/x) then the squeze theorem work by this way since ((a^x −1)/x) is bounced by 0 and −(1/x) and 0→0 and −(1/x)→0 as x→−∞ then ((a^x −1)/x)→0 as x→−∞ in formal way remember that lim_(x→−∞) f(x)=L only if ∀ε>0,∃M∈R,∀x<M,∣f(x)−L∣<ε them lim_(x→−∞) 0⇔M∈(−∞,0),x<M⇒∣0−0∣<ε lim_(x→−∞) −(1/x)=0⇔M≤−(1/ε),x<M x<M<0⇒−ε≤(1/M)<(1/x)<0⇒0<−(1/x)<−(1/M)≤ε ∣−(1/x)−0∣<ε and finaly, choose the same M above lim_(x→−∞) ((a^x −1)/x)=0 x<M 0<((a^x −1)/x)<−(1/x)<−(1/M)≤ε ∣((a^x −1)/x)−0∣<ε as desired :)

$$\mathrm{we}\:\mathrm{have}\:\mathrm{that}\:\mathrm{for}\:{a}>\mathrm{1},\:{a}^{{x}} \:\mathrm{is}\:\mathrm{crescent},\:\mathrm{then} \\ $$$${x}<\mathrm{0}\Rightarrow{a}^{{x}} <{a}^{\mathrm{0}} =\mathrm{1}\Rightarrow{a}^{{x}} <\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{then} \\ $$$${a}^{{x}} −\mathrm{1}<\mathrm{1}−\mathrm{1}=\mathrm{0}\Rightarrow{a}^{{x}} −\mathrm{1}<\mathrm{0}\:\:\left(−\mathrm{1}\:\mathrm{in}\:\mathrm{both}\:\mathrm{side}\right) \\ $$$$\frac{{a}^{{x}} −\mathrm{1}}{{x}}>\mathrm{0}\:\left(\boldsymbol{\div}{x},\:\mathrm{and}\:\mathrm{sonce}\:{x}<\mathrm{0},\:<\:\mathrm{turn}\:\mathrm{into}\:>\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{that}\:{a}^{{x}} >\mathrm{0},\:\mathrm{then} \\ $$$${a}^{{x}} −\mathrm{1}>−\mathrm{1}\:\left(−\mathrm{1}\:\mathrm{in}\:\mathrm{both}\:\mathrm{sides}\right) \\ $$$$\frac{{a}^{{x}} −\mathrm{1}}{{x}}<−\frac{\mathrm{1}}{{x}}\:\left(\boldsymbol{\div}{x},{x}<\mathrm{0}\right) \\ $$$$\mathrm{then} \\ $$$$\mathrm{0}<\frac{{a}^{{x}} −\mathrm{1}}{{x}}<−\frac{\mathrm{1}}{{x}} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{squeze}\:\mathrm{theorem}\:\mathrm{work}\:\mathrm{by}\:\mathrm{this}\:\mathrm{way} \\ $$$$\mathrm{since}\: \\ $$$$\frac{{a}^{{x}} −\mathrm{1}}{{x}}\:\mathrm{is}\:\mathrm{bounced}\:\mathrm{by}\:\mathrm{0}\:\mathrm{and}\:−\frac{\mathrm{1}}{{x}}\:\:\mathrm{and} \\ $$$$\mathrm{0}\rightarrow\mathrm{0}\:\mathrm{and}\:−\frac{\mathrm{1}}{{x}}\rightarrow\mathrm{0}\:\mathrm{as}\:{x}\rightarrow−\infty \\ $$$$\mathrm{then}\:\frac{{a}^{{x}} −\mathrm{1}}{{x}}\rightarrow\mathrm{0}\:\mathrm{as}\:{x}\rightarrow−\infty \\ $$$$\mathrm{in}\:\mathrm{formal}\:\mathrm{way}\:\mathrm{remember}\:\mathrm{that} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{f}\left({x}\right)=\mathrm{L} \\ $$$$\mathrm{only}\:\mathrm{if} \\ $$$$\forall\epsilon>\mathrm{0},\exists\mathrm{M}\in\mathbb{R},\forall{x}<\mathrm{M},\mid{f}\left({x}\right)−\mathrm{L}\mid<\epsilon \\ $$$$\mathrm{them} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\mathrm{0}\Leftrightarrow\mathrm{M}\in\left(−\infty,\mathrm{0}\right),{x}<\mathrm{M}\Rightarrow\mid\mathrm{0}−\mathrm{0}\mid<\epsilon \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:−\frac{\mathrm{1}}{{x}}=\mathrm{0}\Leftrightarrow\mathrm{M}\leqslant−\frac{\mathrm{1}}{\epsilon},{x}<\mathrm{M} \\ $$$${x}<\mathrm{M}<\mathrm{0}\Rightarrow−\epsilon\leqslant\frac{\mathrm{1}}{\mathrm{M}}<\frac{\mathrm{1}}{{x}}<\mathrm{0}\Rightarrow\mathrm{0}<−\frac{\mathrm{1}}{{x}}<−\frac{\mathrm{1}}{\mathrm{M}}\leqslant\epsilon \\ $$$$\mid−\frac{\mathrm{1}}{{x}}−\mathrm{0}\mid<\epsilon \\ $$$$\mathrm{and}\:\mathrm{finaly},\:\mathrm{choose}\:\mathrm{the}\:\mathrm{same}\:\mathrm{M}\:\mathrm{above} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{{a}^{{x}} −\mathrm{1}}{{x}}=\mathrm{0} \\ $$$${x}<\mathrm{M} \\ $$$$\mathrm{0}<\frac{{a}^{{x}} −\mathrm{1}}{{x}}<−\frac{\mathrm{1}}{{x}}<−\frac{\mathrm{1}}{\mathrm{M}}\leqslant\epsilon \\ $$$$\mid\frac{{a}^{{x}} −\mathrm{1}}{{x}}−\mathrm{0}\mid<\epsilon \\ $$$$\left.\mathrm{as}\:\mathrm{desired}\::\right) \\ $$

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

for a>1, a^x is crescent..... what is the meaning of ′crescent′ If a<1 what term will be used?

$$\mathrm{for}\:{a}>\mathrm{1},\:{a}^{{x}} \:\mathrm{is}\:\mathrm{crescent}….. \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{meaning}\:\mathrm{of}\:'\mathrm{crescent}' \\ $$$$\mathrm{If}\:{a}<\mathrm{1}\:\mathrm{what}\:\mathrm{term}\:\mathrm{will}\:\mathrm{be}\:\mathrm{used}? \\ $$$$ \\ $$

Commented by 123456 last updated on 28/Aug/15

a function is crescent if ∀x,y∈D(f) x>y⇒f(x)>f(y) for 0<a<1, a^x is decrescent, or in other words x<y⇒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∈R/Z

$$\mathrm{a}\:\mathrm{function}\:\mathrm{is}\:\mathrm{crescent}\:\mathrm{if}\:\forall{x},{y}\in\mathrm{D}\left({f}\right) \\ $$$${x}>{y}\Rightarrow{f}\left({x}\right)>{f}\left({y}\right) \\ $$$$\mathrm{for}\:\mathrm{0}<{a}<\mathrm{1},\:{a}^{{x}} \:\mathrm{is}\:\mathrm{decrescent},\:\mathrm{or}\:\mathrm{in}\:\mathrm{other} \\ $$$$\mathrm{words} \\ $$$${x}<{y}\Rightarrow{a}^{{x}} >{a}^{{y}} \\ $$$$\mathrm{if}\:{a}=\mathrm{0},{a}^{{x}} \:\mathrm{is}\:\mathrm{not}\:\mathrm{well}\:\mathrm{defined}\:\mathrm{to}\:{x}<\mathrm{0} \\ $$$$\mathrm{if}\:{a}<\mathrm{0},{a}^{{x}} \:\mathrm{is}\:\mathrm{complex}\:\mathrm{and}\:\mathrm{multi}\:\mathrm{valuated}\:\mathrm{at}\:{x}\in\mathbb{R}/\mathbb{Z} \\ $$

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

$$\boldsymbol{\mathrm{Thanks}}! \\ $$

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

((a^x −1)/x)=(a^x −1)×(1/x) lim_(x→−∝) ((a^x −1)/x)=lim_(x→−∝) (a^x −1)×lim_(x→−∝) ((1/x)) =( lim_(x→−∝) a^x −lim_(x→−∝) (1) )×lim_(x→−∝) ((1/x)) Now lim_(x→−∝) a^x =0 for a>1, lim_(x→−∝) (1)=1 and lim_(x→−∝) ((1/x))=0 Hence lim_(x→−∝) ((a^x −1)/x)=(0−1)×0=0 lim_(x→−∝) ((a^x −1)/x)=0

$$\frac{{a}^{{x}} −\mathrm{1}}{{x}}=\left({a}^{{x}} −\mathrm{1}\right)×\frac{\mathrm{1}}{{x}} \\ $$$$\underset{{x}\rightarrow−\propto} {{lim}}\:\frac{{a}^{{x}} −\mathrm{1}}{{x}}=\underset{{x}\rightarrow−\propto} {{lim}}\:\left({a}^{{x}} −\mathrm{1}\right)×\underset{{x}\rightarrow−\propto} {{lim}}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\:\underset{{x}\rightarrow−\propto} {{lim}}\:{a}^{{x}} −\underset{{x}\rightarrow−\propto} {{lim}}\left(\mathrm{1}\right)\:\right)×\underset{{x}\rightarrow−\propto} {{lim}}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$${Now}\:\underset{{x}\rightarrow−\propto} {{lim}a}^{{x}} =\mathrm{0}\:\:{for}\:{a}>\mathrm{1},\:\underset{{x}\rightarrow−\propto} {{lim}}\left(\mathrm{1}\right)=\mathrm{1}\:{and}\:\:\underset{{x}\rightarrow−\propto} {{lim}}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{0} \\ $$$${Hence}\: \\ $$$$\:\:\underset{{x}\rightarrow−\propto} {{lim}}\:\frac{{a}^{{x}} −\mathrm{1}}{{x}}=\left(\mathrm{0}−\mathrm{1}\right)×\mathrm{0}=\mathrm{0} \\ $$$$\:\:\underset{{x}\rightarrow−\propto} {{lim}}\:\frac{{a}^{{x}} −\mathrm{1}}{{x}}=\mathrm{0} \\ $$

Commented by 123456 last updated on 16/Aug/15

$$\left.\mathrm{alright}.\::\right) \\ $$

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