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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 :)
$$ \\ $$$$\frac{{x}^{{n}} −\alpha^{{n}} }{{x}−\alpha}\:=\:\frac{\left({x}−\alpha\right)\left({x}^{{n}−\mathrm{1}} +{x}^{{n}−\mathrm{2}} \alpha+{x}^{{n}−\mathrm{3}} \alpha^{\mathrm{2}} +…+\alpha^{{n}−\mathrm{1}} \right)}{\left({x}−\alpha\right)}\:\overset{{x}\neq\alpha} {=}\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{x}^{{n}−\mathrm{1}−{k}} \alpha^{{k}} \\ $$$${Observe}\:{that}:\:{n}\alpha^{{n}−\mathrm{1}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\alpha^{{n}−\mathrm{1}} \\ $$$$\Rightarrow\frac{{x}^{{n}} −\alpha^{{n}} }{{x}−\alpha}−{n}\alpha^{{n}−\mathrm{1}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{x}^{{n}−\mathrm{1}−{k}} \alpha^{{k}} −\alpha^{{n}−\mathrm{1}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\alpha^{{k}} \left({x}^{{n}−\mathrm{1}−{k}} −\alpha^{{n}−\mathrm{1}−{k}} \right) \\ $$$$ \\ $$$$=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\alpha^{{k}} \left({x}−\alpha\right)\left({x}^{{n}−\mathrm{2}−{k}} +…+\alpha^{{n}−\mathrm{2}−{k}} \right)=\:\left({x}−{a}\right){S}\left({x}\right) \\ $$$${where}\:{S}\left({x}\right)\:{is}\:{the}\:{total}\:{sum}\:{over}\:{the}\:{variable}\:{x} \\ $$$${going}\:{back}\:{to}\:{the}\:{absolute}\:{value}:\:\mid\frac{{x}^{{n}} −\alpha^{{n}} }{{x}−\alpha}−{n}\alpha^{{n}−\mathrm{1}} \mid=\mid{x}−\alpha\mid\mid{S}\left({x}\right)\mid \\ $$$${Now},\:{we}\:{need}\:{to}\:{bound}\:{S}\left({x}\right)\:{under}\:{a}\:{constant} \\ $$$${Suppose}:\:\mid{x}−\alpha\mid<\mathrm{1}\Rightarrow\mid{x}\mid=\mid{x}−\alpha+\alpha\mid\leqslant\mid{x}−\alpha\mid+\mid\alpha\mid<\:\mathrm{1}+\mid\alpha\mid \\ $$$$ \\ $$$${Notice}\:{that}:\:\mid{S}\left({x}\right)\mid=\mid\underset{{terms}} {\sum}{s}_{{k}} \left({x}\right)\mid \\ $$$${where}\:\mid{s}_{{k}} \left({x}\right)\mid=\mid{x}^{{n}−\mathrm{2}−{k}} \alpha^{{k}} \mid=\mid{x}^{{n}−\mathrm{2}−{k}} \mid\mid\alpha^{{k}} \mid\leqslant\mid\alpha^{{k}} \mid\mid\left(\mathrm{1}+\alpha\right)^{{n}−\mathrm{2}−{k}} \mid\:\forall{k} \\ $$$$\Rightarrow{s}\left({k}\right)\:{depends}\:{only}\:{on}\:{n}\:{and}\:\alpha \\ $$$$\therefore\:{s}_{{k}} \left({x}\right)\in\mathbb{R}\:{and}\:{is}\:{finite}\:\forall{k} \\ $$$${choose}\:{K}\in\mathbb{R}\:{s}.{t}.\:\mid{S}\left({x}\right)\mid\leqslant{K} \\ $$$$\therefore\:\mid\frac{{x}^{{n}} −\alpha^{{n}} }{{x}−\alpha}−{n}\alpha^{{n}} \mid\leqslant{K}\mid{x}−\alpha\mid<{K}\delta \\ $$$${Choose}\:\delta={min}\left\{\mathrm{1},\frac{\epsilon}{{K}}\right\}\:\blacksquare \\ $$$$ \\ $$$$\left.{I}\:{hope}\:{I}\:{could}\:{help}\::\right) \\ $$

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