Question

If ${lim}_{x\to 0}\frac{a{e}^x+b\cos x+c{e}^{-x}}{e^{2x}-2e^x+1}=4$, then what are the values of $a,b,c$?

Answer 1

$Given,{lim}_{x\to 0}\frac{a{e}^x+b\cos x+c{e}^{-x}}{e^{2x}-2e^x+1}=4⟶\left(1\right)Whenx\to 0e^{2x}-2e^x+1\to 0\therefore forlimittoexista{e}^x+b\cos x+c{e}^{-x}\to 0Putx=0,a+b+c=0⟶\left(2\right)$

By L'Hospital Rule,differentiating

${lim}_{x\to 0}\frac{a{e}^x-b\sin x-c{e}^{-x}}{{2e}^{2x}-2e^x}=4$

$As,2(e^{2x}-e^x)\to 0\Rightarrow a{e}^x-b\sin x-c{e}^{-x}\to 0$

$Putx=0,a-c=0\Rightarrow a=c⟶\left(3\right)Differentiatingagain$

${lim}_{x\to 0}\frac{a{e}^x-b\cos x+c{e}^{-x}}{{4e}^{2x}-2e^x}=4$

$Putx=0,\frac{a-b+c}{4-2}=4a-b+c=8⟶\left(4\right)Solving\left(2\right),\left(3\right),\left(4\right)a+b+c=0a-b+c=8a=c$

$Adding\left(2\right),\left(4\right)\Rightarrow 2(a+c)=84a=8a=2,c=2Substitutingin\left(2\right),b=-4\therefore a=2,b=-4,c=2$