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lim-x-a-x-a-3-a-a-3-x-a-P-a-lt-0-lim-x-a-x-a-2-a-a-2-x-2-ax-

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Question Number 80650 by jagoll last updated on 05/Feb/20
lim_(x→a) (((∣x∣−a)^3 −(∣a∣−a)^3 )/(x−a)) = P , a <0 lim_(x→a) (((∣x∣−a)^2 −(∣a∣−a)^2 )/(x^2 −ax))=?
$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{\left(\mid{x}\mid−{a}\right)^{\mathrm{3}} −\left(\mid{a}\mid−{a}\right)^{\mathrm{3}} }{{x}−{a}}\:=\:{P}\:,\:{a}\:<\mathrm{0} \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{\left(\mid{x}\mid−{a}\right)^{\mathrm{2}} −\left(\mid{a}\mid−{a}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} −{ax}}=? \\ $$
Commented by john santu last updated on 05/Feb/20
lim_(x→a) (((∣x∣−a)−(∣a∣−a))/(x−a)) = (P/(3(−2a)^2 )) =(P/(12a^2 )) ⇒lim_(x→a) (((∣x∣−a)−(∣a∣−a))/(x−a)) ×lim_(x→a) (((∣x∣−a)+(∣a∣−a))/x)= (P/(12a^2 ))×((−4a)/a)=−(P/(3a^2 ))
$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{\left(\mid{x}\mid−{a}\right)−\left(\mid{a}\mid−{a}\right)}{{x}−{a}}\:=\:\frac{{P}}{\mathrm{3}\left(−\mathrm{2}{a}\right)^{\mathrm{2}} }\:=\frac{{P}}{\mathrm{12}{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{\left(\mid{x}\mid−{a}\right)−\left(\mid{a}\mid−{a}\right)}{{x}−{a}}\:×\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{\left(\mid{x}\mid−{a}\right)+\left(\mid{a}\mid−{a}\right)}{{x}}= \\ $$$$\frac{{P}}{\mathrm{12}{a}^{\mathrm{2}} }×\frac{−\mathrm{4}{a}}{{a}}=−\frac{{P}}{\mathrm{3}{a}^{\mathrm{2}} } \\ $$
Commented by jagoll last updated on 05/Feb/20
thx mister john
$${thx}\:{mister}\:{john} \\ $$
Answered by MJS last updated on 05/Feb/20
P doesn′t exist
$${P}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist} \\ $$
Commented by jagoll last updated on 05/Feb/20
why?
$${why}? \\ $$
Commented by jagoll last updated on 05/Feb/20
i′m correct my question sir
$${i}'{m}\:{correct}\:{my}\:{question}\:{sir} \\ $$
Commented by MJS last updated on 05/Feb/20
for the corrected question P=−12a^2 and the 2^(nd) limit is 4 ∀a<0 which is −(P/(3a^2 ))
$$\mathrm{for}\:\mathrm{the}\:\mathrm{corrected}\:\mathrm{question} \\ $$$${P}=−\mathrm{12}{a}^{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{limit}\:\mathrm{is}\:\mathrm{4}\:\forall{a}<\mathrm{0}\:\mathrm{which}\:\mathrm{is}\:−\frac{{P}}{\mathrm{3}{a}^{\mathrm{2}} } \\ $$
Commented by jagoll last updated on 05/Feb/20
thank you mister
$${thank}\:{you}\:{mister} \\ $$
Answered by MJS last updated on 05/Feb/20
a<0 ⇒ x<0 if x close enough to a (1) ⇒ (((∣x∣−a)^3 −(∣a∣−a)^3 )/(x−a))=−x^2 −4ax−7a^2 =_(x→a) =−12a^2 (2) ⇒ (((∣x∣−a)^2 −(∣a∣−a)^2 )/(x^2 −ax))=((3a)/x)+1=_(x→a) 4
$${a}<\mathrm{0}\:\Rightarrow\:{x}<\mathrm{0}\:\mathrm{if}\:{x}\:\mathrm{close}\:\mathrm{enough}\:\mathrm{to}\:{a} \\ $$$$\left(\mathrm{1}\right)\:\Rightarrow\:\frac{\left(\mid{x}\mid−{a}\right)^{\mathrm{3}} −\left(\mid{a}\mid−{a}\right)^{\mathrm{3}} }{{x}−{a}}=−{x}^{\mathrm{2}} −\mathrm{4}{ax}−\mathrm{7}{a}^{\mathrm{2}} \underset{{x}\rightarrow{a}} {=} \\ $$$$=−\mathrm{12}{a}^{\mathrm{2}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\Rightarrow\:\frac{\left(\mid{x}\mid−{a}\right)^{\mathrm{2}} −\left(\mid{a}\mid−{a}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} −{ax}}=\frac{\mathrm{3}{a}}{{x}}+\mathrm{1}\underset{{x}\rightarrow{a}} {=}\mathrm{4} \\ $$
Commented by jagoll last updated on 05/Feb/20
how to get −x^2 −4ax−7a^2 mister?
$${how}\:{to}\:{get}\:−{x}^{\mathrm{2}} −\mathrm{4}{ax}−\mathrm{7}{a}^{\mathrm{2}} \:{mister}? \\ $$
Commented by MJS last updated on 05/Feb/20
r<0 ⇒ ∣r∣=−r (((−x−a)^3 −(−a−a)^3 )/(x−a))=((−(x+a)^3 +(2a)^3 )/(x−a))= =((−x^3 −3ax^2 −3a^2 x+7a^3 )/(x−a))=−(((x−a)(x^2 +4ax+7a^2 ))/(x−a))= =−x^2 −4ax−7a^2
$${r}<\mathrm{0}\:\Rightarrow\:\mid{r}\mid=−{r} \\ $$$$\frac{\left(−{x}−{a}\right)^{\mathrm{3}} −\left(−{a}−{a}\right)^{\mathrm{3}} }{{x}−{a}}=\frac{−\left({x}+{a}\right)^{\mathrm{3}} +\left(\mathrm{2}{a}\right)^{\mathrm{3}} }{{x}−{a}}= \\ $$$$=\frac{−{x}^{\mathrm{3}} −\mathrm{3}{ax}^{\mathrm{2}} −\mathrm{3}{a}^{\mathrm{2}} {x}+\mathrm{7}{a}^{\mathrm{3}} }{{x}−{a}}=−\frac{\left({x}−{a}\right)\left({x}^{\mathrm{2}} +\mathrm{4}{ax}+\mathrm{7}{a}^{\mathrm{2}} \right)}{{x}−{a}}= \\ $$$$=−{x}^{\mathrm{2}} −\mathrm{4}{ax}−\mathrm{7}{a}^{\mathrm{2}} \\ $$$$ \\ $$
Commented by jagoll last updated on 05/Feb/20
oo i understand sir. thank much
$${oo}\:{i}\:{understand}\:{sir}.\:{thank}\:{much} \\ $$

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