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Question Number 84283 by jagoll last updated on 11/Mar/20

lim_(x→a) (((∣x∣−1)^2 −(∣a∣−1)^2 )/(x−a)) = k , a>0 lim_(x→a) (((∣x∣−1)^3 −(∣a∣−1)^3 )/(x^2 −a^2 )) =

$$\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\left(\mid\mathrm{x}\mid−\mathrm{1}\right)^{\mathrm{2}} −\left(\mid\mathrm{a}\mid−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{x}−\mathrm{a}}\:=\:\mathrm{k}\:,\:\mathrm{a}>\mathrm{0} \\ $$ $$\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\left(\mid\mathrm{x}\mid−\mathrm{1}\right)^{\mathrm{3}} −\left(\mid\mathrm{a}\mid−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }\:=\: \\ $$ $$ \\ $$

Answered by john santu last updated on 11/Mar/20

lim_(x→a) (((x−1)^2 −(a−1)^2 )/(x−a)) = k lim_(x→a) ((x^2 −2x+1−a^2 +2a−1)/(x−a)) = k lim_(x→a) (((x−a)(x+a)−2(x−a))/(x−a)) =k lim_(x→a) (((x−a)(x+a−2))/(x−a)) = k ⇒ k = 2a−2 ⇒ a−1 = (1/2)k (2)lim_(x→a) (((x−1)^3 −(a−1)^3 )/(x^2 −a^2 )) = lim_(x→a) ((3(x−1)^2 )/(2x)) = ((3(a−1)^2 )/(2a)) =((3×(1/4)k^2 )/(2a))= ((3k^2 )/(8a))

$$\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{a}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{x}−\mathrm{a}}\:=\:\mathrm{k} \\ $$ $$\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{1}−\mathrm{a}^{\mathrm{2}} +\mathrm{2a}−\mathrm{1}}{\mathrm{x}−\mathrm{a}}\:=\:\mathrm{k} \\ $$ $$\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}−\mathrm{a}\right)\left(\mathrm{x}+\mathrm{a}\right)−\mathrm{2}\left(\mathrm{x}−\mathrm{a}\right)}{\mathrm{x}−\mathrm{a}}\:=\mathrm{k} \\ $$ $$\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}−\mathrm{a}\right)\left(\mathrm{x}+\mathrm{a}−\mathrm{2}\right)}{\mathrm{x}−\mathrm{a}}\:=\:\mathrm{k} \\ $$ $$\Rightarrow\:\mathrm{k}\:=\:\mathrm{2a}−\mathrm{2}\:\Rightarrow\:\mathrm{a}−\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{k} \\ $$ $$\left(\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} −\left(\mathrm{a}−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }\:=\: \\ $$ $$\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2x}}\:=\:\frac{\mathrm{3}\left(\mathrm{a}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2a}}\: \\ $$ $$=\frac{\mathrm{3}×\frac{\mathrm{1}}{\mathrm{4}}\mathrm{k}^{\mathrm{2}} }{\mathrm{2a}}=\:\frac{\mathrm{3k}^{\mathrm{2}} }{\mathrm{8a}} \\ $$

Commented byjagoll last updated on 11/Mar/20

thank you mister

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mister} \\ $$

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