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Question Number 132328 by liberty last updated on 13/Feb/21
lim_(x→1) ((p−x−x^2 −x^3 −...−x^p )/(x−1)) =?
$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{p}−\mathrm{x}−\mathrm{x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{3}} −…−\mathrm{x}^{\mathrm{p}} }{\mathrm{x}−\mathrm{1}}\:=? \\ $$
Answered by EDWIN88 last updated on 13/Feb/21
The limit is form (0/0) L′Ho^� pital ⇒ L = lim_(x→1) ((−1−2x−3x^2 −4x^3 −...−px^(p−1) )/1) L = −1−2−3−4−...−p L=−(1+2+3+4+...+p)_(AP) = −((p(p+1))/2)
$$\:\mathrm{The}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{form}\:\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\:\mathrm{L}'\mathrm{H}\hat {\mathrm{o}pital}\:\Rightarrow\:\mathrm{L}\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{−\mathrm{1}−\mathrm{2x}−\mathrm{3x}^{\mathrm{2}} −\mathrm{4x}^{\mathrm{3}} −…−\mathrm{px}^{\mathrm{p}−\mathrm{1}} }{\mathrm{1}} \\ $$$$\mathrm{L}\:=\:−\mathrm{1}−\mathrm{2}−\mathrm{3}−\mathrm{4}−…−\mathrm{p} \\ $$$$\mathrm{L}=−\underset{\mathrm{AP}} {\underbrace{\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…+\mathrm{p}\right)}}\:=\:−\frac{\mathrm{p}\left(\mathrm{p}+\mathrm{1}\right)}{\mathrm{2}} \\ $$
Answered by mathmax by abdo last updated on 13/Feb/21
f(x)=((p−(x+x^2 +....+x^p ))/(x−1)) we do the cha7gement x−1=t (sot→o) ⇒f(x)=f(1+t) =((p−(1+t +(1+t)^2 +....(1+t)^p ))/t) ⇒ f(1+t)∼((p−(1+t+1+2t +....+1+pt))/t)=((p−p−(1+2+3...+p)t)/t) =−((p(p+1))/2) ⇒lim_(t→0) f(1+t)=−((p(p+1))/2)=lim_(x→1) f(x)
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{p}−\left(\mathrm{x}+\mathrm{x}^{\mathrm{2}} +….+\mathrm{x}^{\mathrm{p}} \right)}{\mathrm{x}−\mathrm{1}}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{cha7gement}\:\mathrm{x}−\mathrm{1}=\mathrm{t}\:\left(\mathrm{sot}\rightarrow\mathrm{o}\right) \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{1}+\mathrm{t}\right)\:=\frac{\mathrm{p}−\left(\mathrm{1}+\mathrm{t}\:+\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} +….\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{p}} \right)}{\mathrm{t}}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{1}+\mathrm{t}\right)\sim\frac{\mathrm{p}−\left(\mathrm{1}+\mathrm{t}+\mathrm{1}+\mathrm{2t}\:+….+\mathrm{1}+\mathrm{pt}\right)}{\mathrm{t}}=\frac{\mathrm{p}−\mathrm{p}−\left(\mathrm{1}+\mathrm{2}+\mathrm{3}…+\mathrm{p}\right)\mathrm{t}}{\mathrm{t}} \\ $$$$=−\frac{\mathrm{p}\left(\mathrm{p}+\mathrm{1}\right)}{\mathrm{2}}\:\Rightarrow\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{1}+\mathrm{t}\right)=−\frac{\mathrm{p}\left(\mathrm{p}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right) \\ $$

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