[College Calc 01-03. Limits of Functions] #02

Claim: $\lim_{x\rightarrow 0}\frac{e-e^{\cos x}}{x^2}=\frac{e}{2}$ Proof: Since the limit is of the form $\frac{e-e^{\cos 0}}{0^2}=\frac{0}{0} \;,$ L'Hôpital's Rule is applicable. So we have, $\lim_{x\rightarrow 0}\frac{e-e^{\cos x}}{x^2}=\lim_{x\rightarrow 0}\frac{\frac{d}{dx}(e-e^{\cos x})}{\frac{d}{dx}(x^2)}=\lim_{x\rightarrow 0}\frac{\sin x\cdot e^{\cos x}}{2x}=\frac{0}{0}$ Applying L'Hôpital's Rule again, we have $\lim_{x\rightarrow 0}\frac{\sin x\cdot e^{\cos x}}{2x}=\lim_{x\rightarrow 0}\frac{\frac{d}{dx}\sin x\cdot e^{\cos x}}{\frac{d}{dx}(2x)}=\lim_{x\rightarrow 0}\frac{\cos x\cdot e^{\cos x}-\sin^2x\cdot e^{\cos x}}{2}=\frac{\cos 0\cdot e^{\cos 0}-\sin^20\cdot e^{\cos 0}}{2}=\frac{e}{2}$

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Claim: $\lim_{x\rightarrow 0}\; (\cos x)^{\large \frac{1}{x^2}}=\frac{\sqrt{e}}{e}$ Proof: Let $f(x)= (\cos x)^{\large \frac{1}{x^2}}$ . In order to eliminate the exponent, $\log_e f(x)=\log_e (\cos x)^{\large \frac{1}{x^2}}=\frac{1}{x^2}\log_e(\cos x)$ $\implies \lim_{x\rightarrow 0}\log_ef(x)=\lim_{x\rightarrow 0}\frac{\log_e(\cos x)}{x^2}=\frac{0}{0}$ Applying L'Hôpital's Rule, we have $\lim_{x\rightarrow 0}\log_ef(x)=\lim_{x\rightarrow 0}\frac{\frac{d}{dx}\log_e(\cos x)}{\frac{d}{dx}x^2}=\lim_{x\rightarrow 0}\frac{-\tan x}{2x} =-\frac{1}{2}\cdot \lim_{x\rightarrow 0}\frac{\tan x}{x}=-\frac{1}{2}$ $\implies \lim_{x\rightarrow 0}\; (\cos x)^{\large \frac{1}{x^2}}= \lim_{x\rightarrow 0}f(x)=\frac{1}{\sqrt{e}}=\frac{\sqrt{e}}{e}$

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Using the above two claims, from the prioperties of limits , $\lim_{x\rightarrow 0}\frac{(e-e^{\cos x})(\cos x)^{\large \frac{1}{x^2}}}{x^2}=\lim_{x\rightarrow 0}\frac{e-e^{\cos x}}{x^2}\cdot \lim_{x\rightarrow 0}\; (\cos x)^{\large \frac{1}{x^2}}$ $\;\;\;\;\;\;\;=\frac{e}{2}\cdot \frac{\sqrt{e}}{e}$ $\;=\frac{\sqrt{e}}{2}$

Therefore, $p=q=\frac{1}{2}\implies 360(p^2+q)=\boxed{270}$

$\alpha=\lim_{x\rightarrow{0}} \mathrm{\frac{(e-e^{cosx})(cosx)^{\frac{1}{x^2}}}{x^2}}$ $\color{#E81990}\mathrm{\lim_{x\rightarrow{0}}(cosx)^{\frac{1}{x^2}} = (1-{\frac{x^2}{2}})^{\frac{1}{x^2}} = (1-\frac{x^2}{2})^{-\frac{2}{x^2} .-\frac{1}{2}}=e^{-\frac{1}{2}}}$ Note $[\color{#20A900}\mathrm{cosx=1-\frac{x^2}{2!} +\frac{x^4}{4!}-\cdots}]$ and $[\mathrm{\lim_{x\rightarrow{0} } (1+x)^{\frac{1}{x}}=e}]$ $\alpha= \lim_{x\rightarrow{0}}\mathrm{\frac{(e-e^{cosx})e^{-1/2}}{x^2}=\frac{\sqrt{e}-e^{\frac{1-x^2}{2}}}{x^2}=\sqrt{e}{(\frac{1-e^{-x^2/2}}{x^2})}}$ $\mathrm{\lim_{x\rightarrow{0}}\sqrt{e}{(\frac{1-1+\frac{x^2}{2}}{x^2})}=\mathrm{\boxed{\frac{\sqrt{e}}{2}}}}$ Note : Here $\color{#3D99F6} \mathrm{\lim_{x\rightarrow{0}} e^{-x^2/2} ≈ 1-\frac{x^2}{2}}$ and $\color{#3D99F6} \mathrm{\lim_{x\rightarrow{0}} cosx≈1-\frac{x^2}{2}}$

So $p=\frac{1}{2}$ and $q=\frac{1}{2}$

Answer is $\color{#CEBB00}\boxed{360((\frac{1}{2})^2 +\frac{1}{2})= 270}$

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