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Question Number 161533 by mathlove last updated on 19/Dec/21

Commented by cortano last updated on 19/Dec/21

 lim_(x→0)  ((((√(1+x^2 +x)))^(2020) −((√(1+x^2 −x)))^(2020) )/x)   = lim_(x→0)  (((1+1010(x^2 +x))−(1+1010(x^2 −x)))/x)   = lim_(x→0)  ((2020x)/x) = 2020

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} +{x}}\right)^{\mathrm{2020}} −\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} −{x}}\right)^{\mathrm{2020}} }{{x}} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{1010}\left({x}^{\mathrm{2}} +{x}\right)\right)−\left(\mathrm{1}+\mathrm{1010}\left({x}^{\mathrm{2}} −{x}\right)\right)}{{x}} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2020}{x}}{{x}}\:=\:\mathrm{2020} \\ $$

Answered by mathmax by abdo last updated on 19/Dec/21

f(x)=(((1+x^2 +x)^(1010) −(1+x^2 −x)^(1010) )/x) ⇒  f(x)∼((1+1010(x^2 +x)−(1+1010(x^2 −x)))/x)  f(x)∼((1010x^2 +1010x−1010x^2 +1010x)/x) ⇒  f(x)∼((2020x)/x) ⇒lim_(x→0) f(x)=2020

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}\right)^{\mathrm{1010}} −\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} −\mathrm{x}\right)^{\mathrm{1010}} }{\mathrm{x}}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{1}+\mathrm{1010}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}\right)−\left(\mathrm{1}+\mathrm{1010}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}\right)\right)}{\mathrm{x}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{1010x}^{\mathrm{2}} +\mathrm{1010x}−\mathrm{1010x}^{\mathrm{2}} +\mathrm{1010x}}{\mathrm{x}}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{2020x}}{\mathrm{x}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{2020} \\ $$

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