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Question Number 128821 by benjo_mathlover last updated on 10/Jan/21
 Nice limit !   For −1<a <1 the value of lim_(x→−a)  (((x^2 +2ax+a^2 )(x^2 +ax+a^2 ))/(((√(1−x^2 ))−(√(1−a^2 )) )^2 )) =?
$$\:\mathrm{Nice}\:\mathrm{limit}\:! \\ $$$$\:\mathrm{For}\:−\mathrm{1}<{a}\:<\mathrm{1}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\underset{{x}\rightarrow−{a}} {\mathrm{lim}}\:\frac{\left({x}^{\mathrm{2}} +\mathrm{2}{ax}+{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{ax}+{a}^{\mathrm{2}} \right)}{\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\:\right)^{\mathrm{2}} }\:=? \\ $$
Answered by liberty last updated on 10/Jan/21
 lim_(x→−a)  (((x+a)^2 (a^2 −a^2 +a^2 ))/(((√(1−x^2 ))−(√(1−a^2 )) )^2 )) = a^2  lim_(x→−a) ((((x+a)((√(1−x^2 ))+(√(1−a^2 ))))/(a^2 −x^2 )))^2   = a^2  lim_(x→−a)  ((((√(1−x^2 )) +(√(1−a^2 )))/(a−x)))^2    = a^2  × (((2(√(1−a^2 )))/(2a)) )^2 = 1−a^2
$$\:\underset{{x}\rightarrow−{a}} {\mathrm{lim}}\:\frac{\left({x}+{a}\right)^{\mathrm{2}} \left({a}^{\mathrm{2}} −{a}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}{\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\:\right)^{\mathrm{2}} }\:=\:{a}^{\mathrm{2}} \:\underset{{x}\rightarrow−{a}} {\mathrm{lim}}\left(\frac{\left({x}+{a}\right)\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\right)}{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$=\:{a}^{\mathrm{2}} \:\underset{{x}\rightarrow−{a}} {\mathrm{lim}}\:\left(\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{{a}−{x}}\right)^{\mathrm{2}} \\ $$$$\:=\:{a}^{\mathrm{2}} \:×\:\left(\frac{\mathrm{2}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{\mathrm{2}{a}}\:\right)^{\mathrm{2}} =\:\mathrm{1}−{a}^{\mathrm{2}} \\ $$

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