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Question Number 227509 by Lara2440 last updated on 06/Feb/26
PLS HELP!!!!! we want to prove lim_(h→0^± ) (((x+h)^n −x^n )/h)=n∙x^(n−1) by using the 𝛆-𝛅 argument def. ∀𝛆>0 , ∃𝛅>0 s.t. ∣x−α∣<𝛅 ⇒∣f(x)−L∣<𝛆 Now, we want to show that for all 𝛆>0 , Exist 𝛅>0 s.t. ∣x−α∣<𝛅 Implies ∣((x^n −α^n )/(x−α))−n∙α^(n−1) ∣<𝛆 But i having a trouble solving this inequality ∣((x^n −α^n )/(x−α))−n∙α^(n−1) ∣<𝛆..... our purpose is: to simplify the Expression ∣((x^n −α^n )/(x−α))−n∙α^(n−1) ∣into the form of K∙∣x−α∣ (K is Constant) and find 𝛅=(1/K)𝛆
$$\boldsymbol{\mathrm{PLS}}\:\boldsymbol{\mathrm{HELP}}!!!!! \\ $$$$\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{prove}\:\underset{{h}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\:\frac{\left({x}+{h}\right)^{{n}} −{x}^{{n}} }{{h}}={n}\centerdot{x}^{{n}−\mathrm{1}} \:\mathrm{by}\:\mathrm{using}\:\mathrm{the}\:\boldsymbol{\epsilon}-\boldsymbol{\delta}\:\mathrm{argument} \\ $$$$\mathrm{def}.\:\forall\boldsymbol{\epsilon}>\mathrm{0}\:,\:\exists\boldsymbol{\delta}>\mathrm{0}\:\mathrm{s}.\mathrm{t}.\:\mid{x}−\alpha\mid<\boldsymbol{\delta}\:\Rightarrow\mid{f}\left({x}\right)−{L}\mid<\boldsymbol{\epsilon} \\ $$$$\mathrm{Now},\:\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\mathrm{for}\:\mathrm{all}\:\boldsymbol{\epsilon}>\mathrm{0}\:,\:\mathrm{Exist}\:\boldsymbol{\delta}>\mathrm{0}\:\mathrm{s}.\mathrm{t}.\:\mid{x}−\alpha\mid<\boldsymbol{\delta}\:\:\mathrm{Implies}\:\mid\frac{{x}^{{n}} −\alpha^{{n}} }{{x}−\alpha}−{n}\centerdot\alpha^{{n}−\mathrm{1}} \mid<\boldsymbol{\epsilon} \\ $$$$\mathrm{But}\:\mathrm{i}\:\mathrm{having}\:\mathrm{a}\:\mathrm{trouble}\:\mathrm{solving}\:\mathrm{this}\:\:\mathrm{inequality}\:\:\mid\frac{{x}^{{n}} −\alpha^{{n}} }{{x}−\alpha}−{n}\centerdot\alpha^{{n}−\mathrm{1}} \mid<\boldsymbol{\epsilon}….. \\ $$$$\mathrm{our}\:\mathrm{purpose}\:\mathrm{is}: \\ $$$$\mathrm{to}\:\mathrm{simplify}\:\mathrm{the}\:\mathrm{Expression}\:\mid\frac{{x}^{{n}} −\alpha^{{n}} }{{x}−\alpha}−{n}\centerdot\alpha^{{n}−\mathrm{1}} \mid\mathrm{into}\:\mathrm{the}\:\mathrm{form}\:\mathrm{of} \\ $$$${K}\centerdot\mid{x}−\alpha\mid\:\left({K}\:\mathrm{is}\:\mathrm{Constant}\right) \\ $$$$\mathrm{and}\:\mathrm{find}\:\boldsymbol{\delta}=\frac{\mathrm{1}}{{K}}\boldsymbol{\epsilon} \\ $$
Commented by fantastic2 last updated on 06/Feb/26
anyone please help lara
$${anyone}\:{please}\:{help}\:{lara} \\ $$
Commented by Lara2440 last updated on 06/Feb/26
yeah
$$\mathrm{yeah}\: \\ $$$$ \\ $$
Answered by Kassista last updated on 06/Feb/26
((x^n −α^n )/(x−α)) = (((x−α)(x^(n−1) +x^(n−2) α+x^(n−3) α^2 +...+α^(n−1) ))/((x−α))) =^(x≠α) Σ_(k=0) ^(n−1) x^(n−1−k) α^k Observe that: nα^(n−1) =Σ_(k=0) ^(n−1) α^(n−1) ⇒((x^n −α^n )/(x−α))−nα^(n−1) =Σ_(k=0) ^(n−1) x^(n−1−k) α^k −α^(n−1) = Σ_(k=0) ^(n−1) α^k (x^(n−1−k) −α^(n−1−k) ) = Σ_(k=0) ^(n−1) α^k (x−α)(x^(n−2−k) +...+α^(n−2−k) )= (x−a)S(x) where S(x) is the total sum over the variable x going back to the absolute value: ∣((x^n −α^n )/(x−α))−nα^(n−1) ∣=∣x−α∣∣S(x)∣ Now, we need to bound S(x) under a constant Suppose: ∣x−α∣<1⇒∣x∣=∣x−α+α∣≤∣x−α∣+∣α∣< 1+∣α∣ Notice that: ∣S(x)∣=∣Σ_(terms) s_k (x)∣ where ∣s_k (x)∣=∣x^(n−2−k) α^k ∣=∣x^(n−2−k) ∣∣α^k ∣≤∣α^k ∣∣(1+α)^(n−2−k) ∣ ∀k ⇒s(k) depends only on n and α ∴ s_k (x)∈R and is finite ∀k choose K∈R s.t. ∣S(x)∣≤K ∴ ∣((x^n −α^n )/(x−α))−nα^n ∣≤K∣x−α∣<Kδ Choose δ=min{1,(ε/K)} ■ I hope I could help :)
$$ \\ $$$$\frac{{x}^{{n}} −\alpha^{{n}} }{{x}−\alpha}\:=\:\frac{\left({x}−\alpha\right)\left({x}^{{n}−\mathrm{1}} +{x}^{{n}−\mathrm{2}} \alpha+{x}^{{n}−\mathrm{3}} \alpha^{\mathrm{2}} +…+\alpha^{{n}−\mathrm{1}} \right)}{\left({x}−\alpha\right)}\:\overset{{x}\neq\alpha} {=}\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{x}^{{n}−\mathrm{1}−{k}} \alpha^{{k}} \\ $$$${Observe}\:{that}:\:{n}\alpha^{{n}−\mathrm{1}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\alpha^{{n}−\mathrm{1}} \\ $$$$\Rightarrow\frac{{x}^{{n}} −\alpha^{{n}} }{{x}−\alpha}−{n}\alpha^{{n}−\mathrm{1}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{x}^{{n}−\mathrm{1}−{k}} \alpha^{{k}} −\alpha^{{n}−\mathrm{1}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\alpha^{{k}} \left({x}^{{n}−\mathrm{1}−{k}} −\alpha^{{n}−\mathrm{1}−{k}} \right) \\ $$$$ \\ $$$$=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\alpha^{{k}} \left({x}−\alpha\right)\left({x}^{{n}−\mathrm{2}−{k}} +…+\alpha^{{n}−\mathrm{2}−{k}} \right)=\:\left({x}−{a}\right){S}\left({x}\right) \\ $$$${where}\:{S}\left({x}\right)\:{is}\:{the}\:{total}\:{sum}\:{over}\:{the}\:{variable}\:{x} \\ $$$${going}\:{back}\:{to}\:{the}\:{absolute}\:{value}:\:\mid\frac{{x}^{{n}} −\alpha^{{n}} }{{x}−\alpha}−{n}\alpha^{{n}−\mathrm{1}} \mid=\mid{x}−\alpha\mid\mid{S}\left({x}\right)\mid \\ $$$${Now},\:{we}\:{need}\:{to}\:{bound}\:{S}\left({x}\right)\:{under}\:{a}\:{constant} \\ $$$${Suppose}:\:\mid{x}−\alpha\mid<\mathrm{1}\Rightarrow\mid{x}\mid=\mid{x}−\alpha+\alpha\mid\leqslant\mid{x}−\alpha\mid+\mid\alpha\mid<\:\mathrm{1}+\mid\alpha\mid \\ $$$$ \\ $$$${Notice}\:{that}:\:\mid{S}\left({x}\right)\mid=\mid\underset{{terms}} {\sum}{s}_{{k}} \left({x}\right)\mid \\ $$$${where}\:\mid{s}_{{k}} \left({x}\right)\mid=\mid{x}^{{n}−\mathrm{2}−{k}} \alpha^{{k}} \mid=\mid{x}^{{n}−\mathrm{2}−{k}} \mid\mid\alpha^{{k}} \mid\leqslant\mid\alpha^{{k}} \mid\mid\left(\mathrm{1}+\alpha\right)^{{n}−\mathrm{2}−{k}} \mid\:\forall{k} \\ $$$$\Rightarrow{s}\left({k}\right)\:{depends}\:{only}\:{on}\:{n}\:{and}\:\alpha \\ $$$$\therefore\:{s}_{{k}} \left({x}\right)\in\mathbb{R}\:{and}\:{is}\:{finite}\:\forall{k} \\ $$$${choose}\:{K}\in\mathbb{R}\:{s}.{t}.\:\mid{S}\left({x}\right)\mid\leqslant{K} \\ $$$$\therefore\:\mid\frac{{x}^{{n}} −\alpha^{{n}} }{{x}−\alpha}−{n}\alpha^{{n}} \mid\leqslant{K}\mid{x}−\alpha\mid<{K}\delta \\ $$$${Choose}\:\delta={min}\left\{\mathrm{1},\frac{\epsilon}{{K}}\right\}\:\blacksquare \\ $$$$ \\ $$$$\left.{I}\:{hope}\:{I}\:{could}\:{help}\::\right) \\ $$
Commented by Lara2440 last updated on 07/Feb/26
thx....... but i understand ∣((x^n −a^n )/(x−a))−na^(n−1) ∣<ε ⇒∣(((x−a)(x^(n−1) +ax^(n−2) +...a^(n−2) x+a^(n−1) ))/((x−a)))−na^(n−1) ∣<ε ∣Σ_(h=1) ^n a^(h−1) x^(n−h) −na^(n−1) ∣<ε but ...why... it bounded???
$$\mathrm{thx}……. \\ $$$$\mathrm{but}\:\mathrm{i}\:\mathrm{understand} \\ $$$$\mid\frac{{x}^{{n}} −{a}^{{n}} }{{x}−{a}}−{na}^{{n}−\mathrm{1}} \mid<\epsilon\:\Rightarrow\mid\frac{\left({x}−{a}\right)\left({x}^{{n}−\mathrm{1}} +{ax}^{{n}−\mathrm{2}} +…{a}^{{n}−\mathrm{2}} {x}+{a}^{{n}−\mathrm{1}} \right)}{\left({x}−{a}\right)}−{na}^{{n}−\mathrm{1}} \mid<\epsilon \\ $$$$\mid\underset{{h}=\mathrm{1}} {\overset{{n}} {\sum}}\:{a}^{{h}−\mathrm{1}} {x}^{{n}−{h}} −{na}^{{n}−\mathrm{1}} \mid<\epsilon\: \\ $$$$\mathrm{but}\:…\mathrm{why}…\:\mathrm{it}\:\mathrm{bounded}??? \\ $$
Commented by Kassista last updated on 07/Feb/26
when i supposed that ∣x−α∣<1 and showed that ,under this condition, ∣x∣≤1+∣α∣ I could replace that into the blablabla and get a representation of each term of S(x) using only α and n, therefore, each term must be a number when you add all this numbers together, you get a finite value for the sum the key is to independ it from the variable x
$${when}\:{i}\:{supposed}\:{that}\:\mid{x}−\alpha\mid<\mathrm{1}\:{and}\:{showed}\:{that} \\ $$$$,{under}\:{this}\:{condition},\:\mid{x}\mid\leqslant\mathrm{1}+\mid\alpha\mid \\ $$$$ \\ $$$${I}\:{could}\:{replace}\:{that}\:{into}\:{the}\:{blablabla}\:{and}\:{get} \\ $$$${a}\:{representation}\:{of}\:{each}\:{term}\:{of}\:{S}\left({x}\right)\:{using}\:{only} \\ $$$$\alpha\:{and}\:{n},\:{therefore},\:{each}\:{term}\:{must}\:{be}\:{a}\:{number} \\ $$$${when}\:{you}\:{add}\:{all}\:{this}\:{numbers}\:{together},\: \\ $$$${you}\:{get}\:{a}\:{finite}\:{value}\:{for}\:{the}\:{sum} \\ $$$$ \\ $$$${the}\:{key}\:{is}\:{to}\:{independ}\:{it}\:{from}\:{the}\:{variable}\:{x} \\ $$
Commented by Lara2440 last updated on 07/Feb/26
So...you′re saying that you use triangle inequality to limit it to an upper bound situation on purpose and then make the variable x a constant so that it′s easier to handle right?
$$\mathrm{So}…\mathrm{you}'\mathrm{re}\:\mathrm{saying}\:\mathrm{that}\:\mathrm{you}\:\mathrm{use}\:\mathrm{triangle}\:\mathrm{inequality}\:\mathrm{to} \\ $$$$\mathrm{limit}\:\mathrm{it}\:\mathrm{to}\:\mathrm{an}\:\mathrm{upper}\:\mathrm{bound}\:\mathrm{situation}\:\mathrm{on}\:\mathrm{purpose} \\ $$$$\mathrm{and}\:\mathrm{then}\:\mathrm{make}\:\mathrm{the}\:\mathrm{variable}\:{x}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{so}\:\mathrm{that} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{easier}\:\mathrm{to}\:\mathrm{handle}\:\mathrm{right}? \\ $$
Commented by Kassista last updated on 07/Feb/26
I think that what I used was the normal triangle inequality, but yes, you′re correct
$${I}\:{think}\:{that}\:{what}\:{I}\:{used}\:{was}\:{the}\:{normal} \\ $$$${triangle}\:{inequality},\:{but}\:{yes},\:{you}'{re}\:{correct} \\ $$
Answered by vnm last updated on 07/Feb/26
(((x+δ)^n −x^n )/δ)=((Σ_(k=0) ^n C_n ^k x^(n−k) δ^k −x^n )/δ)=((Σ_(k=1) ^n C_n ^k x^(n−k) δ^k )/δ)= C_n ^1 x^(n−1) +Σ_(k=2) ^n C_n ^k x^(n−k) δ^(k−1) =nx^(n−1) +Σ_(k=2) ^n C_n ^k x^(n−k) δ^(k−1) if x=0 then Σ_(k=2) ^n C_n ^k x^(n−k) δ^(k−1) =0 else ∣Σ_(k=2) ^n C_n ^k x^(n−k) δ^(k−1) ∣≤C_n ^([(n/2)]) ∣x∣^(n−1) Σ_(k=2) ^n ∣(δ/x)∣^(k−1) C_n ^([(n/2)]) ∣x∣^(n−1) =A doesn′t depend on δ if ∣δ∣<∣x∣ then ∣(δ/x)∣^(k−1) ≤^(k≥2) ∣(δ/x)∣and Σ_(k=2) ^n ∣(δ/x)∣^k ≤(n−1)∣(δ/x)∣ A(n−1)∣(δ/x)∣<ε ∣δ∣<∣x∣∙min((ε/(A(n−1))), 1)
$$\frac{\left({x}+\delta\right)^{{n}} −{x}^{{n}} }{\delta}=\frac{\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} {x}^{{n}−{k}} \delta^{{k}} −{x}^{{n}} }{\delta}=\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} {x}^{{n}−{k}} \delta^{{k}} }{\delta}= \\ $$$${C}_{{n}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} +\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} {x}^{{n}−{k}} \delta^{{k}−\mathrm{1}} ={nx}^{{n}−\mathrm{1}} +\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} {x}^{{n}−{k}} \delta^{{k}−\mathrm{1}} \\ $$$$\mathrm{if}\:{x}=\mathrm{0}\:\mathrm{then}\:\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} {x}^{{n}−{k}} \delta^{{k}−\mathrm{1}} =\mathrm{0} \\ $$$$\mathrm{else} \\ $$$$\mid\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} {x}^{{n}−{k}} \delta^{{k}−\mathrm{1}} \mid\leqslant{C}_{{n}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \mid{x}\mid^{{n}−\mathrm{1}} \underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\mid\frac{\delta}{{x}}\mid^{{k}−\mathrm{1}} \\ $$$${C}_{{n}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \mid{x}\mid^{{n}−\mathrm{1}} \:={A}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{depend}\:\mathrm{on}\:\delta\: \\ $$$$\mathrm{if}\:\mid\delta\mid<\mid{x}\mid\:\mathrm{then}\:\mid\frac{\delta}{{x}}\mid^{{k}−\mathrm{1}} \:\overset{{k}\geqslant\mathrm{2}} {\leqslant}\:\mid\frac{\delta}{{x}}\mid\mathrm{and}\:\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\mid\frac{\delta}{{x}}\mid^{{k}} \leqslant\left({n}−\mathrm{1}\right)\mid\frac{\delta}{{x}}\mid \\ $$$${A}\left({n}−\mathrm{1}\right)\mid\frac{\delta}{{x}}\mid<\varepsilon \\ $$$$\mid\delta\mid<\mid{x}\mid\centerdot\mathrm{min}\left(\frac{\varepsilon}{{A}\left({n}−\mathrm{1}\right)},\:\mathrm{1}\right) \\ $$$$ \\ $$

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