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Question Number 117416 by Lordose last updated on 11/Oct/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{\mathrm{e}}^{{\mathrm{x}}^2}-\mathrm{cosx}}{{\sin }^2\mathrm{x}}$$

Answered by Olaf last updated on 11/Oct/20

$$\mathrm{\forall }x\in \mathbb{R},\sin x⩽x\Rightarrow \frac{1}{{\sin }^{2}x}⩾\frac{1}{x^2}$$$$e^{x^2}-\cos x⩾e^{x^2}-1$$$$\frac{e^{x^2}-\cos x}{{\sin }^{2}x}⩾\frac{e^{x^2}-1}{x^2}$$$$\mathrm{But}\underset{x\to \mathrm{\infty }}{\lim }\frac{e^{x^2}-1}{x^2}=\underset{u\to \mathrm{\infty }}{\lim }\frac{e^u-1}{u}=\underset{u\to \mathrm{\infty }}{\lim }\frac{e^u-0}{1}=+\mathrm{\infty }$$$$\left({\mathrm{Hospital}}^{\prime }\mathrm{s}\mathrm{rule}\right)$$$$\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }\frac{e^{x^2}-\cos x}{{\sin }^{2}x}=+\mathrm{\infty }$$$$\left(\mathrm{by}\mathrm{comparison}\right)$$

Commented by Lordose last updated on 11/Oct/20

$$\mathrm{the}\mathrm{text}\mathrm{said}\frac{3}{2}$$

Commented by MJS_new last updated on 11/Oct/20

$$\mathrm{obviously}\mathrm{the}\mathrm{text}\mathrm{is}\mathrm{wrong}$$

Commented by Olaf last updated on 11/Oct/20

$$dox=n\pi$$$$wehave\frac{e^{{\pi }^2{n}^2}-{(-1)}^n}{0^{+}}\to \mathrm{\infty }$$

Commented by Olaf last updated on 11/Oct/20

$$\frac{3}{2}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{in}0$$

Answered by MJS_new last updated on 11/Oct/20

$$-1⩽\cos x⩽1$$$$0⩽{\sin }^{2}x⩽1$$$$\Rightarrow \lim =\mathrm{\infty }$$

Answered by Bird last updated on 12/Oct/20

$$letf\left(x\right)=\frac{e^{x^2}-cosx}{{sin}^{2}x}\Rightarrow$$$$f\left(x\right)=2\frac{e^{x^2}-cosx}{1-cos\left(2x\right)}$$$$wehavecosu=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{u}^{2n}}{\left(2n\right)!}$$$$=1-\frac{u^2}{2}+\frac{u^4}{4!}-..\Rightarrow 1-\frac{u^2}{2}⩽cosu⩽1-\frac{u^2}{2}+\frac{u^4}{4!}$$$$-1+\frac{u^2}{2}-\frac{u^4}{4!}⩽-cosu⩽-1+\frac{u^2}{2}$$$$\Rightarrow \frac{u^2}{2}-\frac{u^4}{4!}⩽1-cosu⩽\frac{u^2}{2}\Rightarrow$$$$\frac{2}{u^2}⩽\frac{1}{1-cosu}⩽\frac{1}{\frac{u^2}{2}-\frac{u^4}{4!}}\Rightarrow$$$$\frac{1}{2x^2}⩽\frac{1}{1-cos\left(2x\right)}⩽\frac{1}{2x^2-\frac{16}{4!}{x}^4}\Rightarrow$$$$\frac{e^{x^2}-cosx}{2x^2}⩽\frac{e^{x^2}-cosx}{1-cosx}$$$$but{lim}_{x\to +\mathrm{\infty }}\frac{e^{x^2}-cosx}{2x^2}$$$$={lim}_{x\to +\mathrm{\infty }}\frac{e^{x^2}}{2x^2}=+\mathrm{\infty }\Rightarrow$$$${lim}_{x\to +\mathrm{\infty }}f\left(x\right)=+\mathrm{\infty }$$$$$$

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