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Question Number 128908 by bramlexs22 last updated on 11/Jan/21

 lim_(x→∞)  (x(√(x^2 +4)) −(√(x^4 +16)) )=?

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{x}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}\:−\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{16}}\:\right)=? \\ $$

Answered by Dwaipayan Shikari last updated on 11/Jan/21

(x^2 (√(1+(4/x^2 )))−x^2 (√(1+((16)/x^4 ))))=x^2 (1+(2/x^2 )−1−(8/x^4 ))=2

$$\left({x}^{\mathrm{2}} \sqrt{\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }}−{x}^{\mathrm{2}} \sqrt{\mathrm{1}+\frac{\mathrm{16}}{{x}^{\mathrm{4}} }}\right)={x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }−\mathrm{1}−\frac{\mathrm{8}}{{x}^{\mathrm{4}} }\right)=\mathrm{2} \\ $$

Answered by liberty last updated on 11/Jan/21

standard way    lim_(x→∞)  ((((√(x^4 +4x^2 )) −(√(x^4 +16)))((√(x^4 +4x^2 ))+(√(x^4 +16))))/( (√(x^4 +4x^2 ))+(√(x^4 +16)))) =   lim_(x→∞)  ((4x^2 −16)/(x^2  ((√(1+4x^(−2) ))+(√(1+16x^(−4) )) ))) =    lim_(x→∞)  ((4−16x^(−2) )/( (√(1+4x^(−2) ))+(√(1+16x^(−4) )))) = (4/2)=2

$$\mathrm{standard}\:\mathrm{way}\: \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{4x}^{\mathrm{2}} }\:−\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{16}}\right)\left(\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{4x}^{\mathrm{2}} }+\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{16}}\right)}{\:\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{4x}^{\mathrm{2}} }+\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{16}}}\:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{4x}^{\mathrm{2}} −\mathrm{16}}{\mathrm{x}^{\mathrm{2}} \:\left(\sqrt{\mathrm{1}+\mathrm{4x}^{−\mathrm{2}} }+\sqrt{\mathrm{1}+\mathrm{16x}^{−\mathrm{4}} }\:\right)}\:=\: \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{4}−\mathrm{16x}^{−\mathrm{2}} }{\:\sqrt{\mathrm{1}+\mathrm{4x}^{−\mathrm{2}} }+\sqrt{\mathrm{1}+\mathrm{16x}^{−\mathrm{4}} }}\:=\:\frac{\mathrm{4}}{\mathrm{2}}=\mathrm{2} \\ $$

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