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lim-x-0-x-2-1-x-2-1-2000x-5-2000x-5-




Question Number 128624 by john_santu last updated on 09/Jan/21
 lim_(x→0) ((∣x^2 +1∣−∣x^2 −1∣)/(∣2000x+5∣−∣2000x−5∣)) =?
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mid\mathrm{x}^{\mathrm{2}} +\mathrm{1}\mid−\mid\mathrm{x}^{\mathrm{2}} −\mathrm{1}\mid}{\mid\mathrm{2000x}+\mathrm{5}\mid−\mid\mathrm{2000x}−\mathrm{5}\mid}\:=? \\ $$
Answered by liberty last updated on 09/Jan/21
 lim_(x→0) (((x^2 +1)+(x^2 −1))/((2000x+5)+(2000x−5)))=lim_(x→0)  ((2x^2 )/(4000x)) = 0
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)+\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\mathrm{2000x}+\mathrm{5}\right)+\left(\mathrm{2000x}−\mathrm{5}\right)}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2x}^{\mathrm{2}} }{\mathrm{4000x}}\:=\:\mathrm{0} \\ $$
Answered by mathmax by abdo last updated on 09/Jan/21
lim_(x→0_+ )    f(x)=lim_(x→0^+ )     ((x^2 +1−(1−x^2 ))/(2000x +5−2000x+5)) =lim_(x→0^+ )   ((2x^2 )/(10))=0  lim_(x→0^− )   f(x)=lim_(x→0^− )    ((x^2 +1−(1−x^2 ))/(5−2000x−(2000x+5)))=lim_(x→0^− )   ((2x^2 )/(−4000x))  =−lim_(x→0^− )   (x/(2000))=0
$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}_{+} } \:\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{1}−\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{2000x}\:+\mathrm{5}−\mathrm{2000x}+\mathrm{5}}\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}^{+} } \:\:\frac{\mathrm{2x}^{\mathrm{2}} }{\mathrm{10}}=\mathrm{0} \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}^{−} } \:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}^{−} } \:\:\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{1}−\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{5}−\mathrm{2000x}−\left(\mathrm{2000x}+\mathrm{5}\right)}=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}^{−} } \:\:\frac{\mathrm{2x}^{\mathrm{2}} }{−\mathrm{4000x}} \\ $$$$=−\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}^{−} } \:\:\frac{\mathrm{x}}{\mathrm{2000}}=\mathrm{0} \\ $$

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