Menu Close

lim-x-0-e-x-e-x-2-x-2-let-x-2t-x-0-t-0-lim-x-0-e-x-e-x-2-x-2-lim-t-0-e-2t-e-2t-2-4t-2-1-4-lim-t-0-e-t-e-t-t-2-1-4-lim-t-0-




Question Number 184903 by aba last updated on 13/Jan/23
lim_(x→0) ((e^x +e^(−x) −2)/x^2 )=??  let x=2t  { ((x→0)),((t→0)) :}  lim_(x→0) ((e^x +e^(−x) −2)/x^2 )=lim_(t→0) ((e^(2t) +e^(−2t) −2)/(4t^2 ))=(1/4)lim_(t→0) (((e^t −e^(−t) )/t))^2 =(1/4)lim_(t→0) (((e^t −1)/t)+((e^(−t) −1)/(−t)))^2 =(1/4)(1+1)^2 =1
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} −\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }=?? \\ $$$$\mathrm{let}\:\mathrm{x}=\mathrm{2t}\:\begin{cases}{\mathrm{x}\rightarrow\mathrm{0}}\\{\mathrm{t}\rightarrow\mathrm{0}}\end{cases} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} −\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{2t}} +\mathrm{e}^{−\mathrm{2t}} −\mathrm{2}}{\mathrm{4t}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} }{\mathrm{t}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{e}^{\mathrm{t}} −\mathrm{1}}{\mathrm{t}}+\frac{\mathrm{e}^{−\mathrm{t}} −\mathrm{1}}{−\mathrm{t}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$
Answered by manxsol last updated on 13/Jan/23
  1+(x/(1!))+(x^2 /(2!))+(x^3 /(3!))+1−(x/(1!))+(x^2 /(2!))−(x^3 /(3!_ ))−2=((2(x^2 /(2!))+2(x^4 /(4!)))/x^2 )=1+(x^2 /(12)).......  lim_(x→0) 1+(x^2 /(12))+.....=1
$$ \\ $$$$\mathrm{1}+\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\mathrm{1}−\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!_{} }−\mathrm{2}=\frac{\mathrm{2}\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\mathrm{2}\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}}{{x}^{\mathrm{2}} }=\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{12}}……. \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{12}}+…..=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by aba last updated on 13/Jan/23
nice
$$\mathrm{nice} \\ $$
Commented by manxsol last updated on 14/Jan/23
  ((e^(0.0000000001) +e^(−0.0000000001) −2)/((0.0000000001)^2 ))  0.0  result of servidor  ((e^(0.00001) +e^(−0.00001) −2)/(0.00001^2 ))  1.0  result of servidor
$$ \\ $$$$\frac{{e}^{\mathrm{0}.\mathrm{0000000001}} +{e}^{−\mathrm{0}.\mathrm{0000000001}} −\mathrm{2}}{\left(\mathrm{0}.\mathrm{0000000001}\right)^{\mathrm{2}} } \\ $$$$\mathrm{0}.\mathrm{0} \\ $$$${result}\:{of}\:{servidor} \\ $$$$\frac{{e}^{\mathrm{0}.\mathrm{00001}} +{e}^{−\mathrm{0}.\mathrm{00001}} −\mathrm{2}}{\mathrm{0}.\mathrm{00001}^{\mathrm{2}} } \\ $$$$\mathrm{1}.\mathrm{0} \\ $$$${result}\:{of}\:{servidor} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *