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Question Number 72251 by aliesam last updated on 26/Oct/19

Commented by mathmax by abdo last updated on 26/Oct/19

$$letf\left(x\right)=\frac{ln(1+ax+{bx}^2)-ln(1-ax+{bx}^2)}{1-cosx}$$$$wehave{ln}^{{}^{\prime }}(1+u)=\frac{1}{1+u}=1-u+o\left(u\right)(u\to 0)\Rightarrow$$$$ln(1+u)=u-\frac{u^2}{2}+o\left(u^2\right)\Rightarrow ln(1+ax+{bx}^2)\sim (ax+{bx}^2)-\frac{{(ax+{bx}^2)}^2}{2}$$$$=ax+{bx}^2-\frac{1}{2}(a^2{x}^2+2{abx}^3+b^2{x}^4)also$$$$ln(1-ax+{bx}^2)\sim -ax+{bx}^2-\frac{1}{2}(a^2{x}^2-2{abx}^3+b^2{x}^4)(changexby-x)$$$$\Rightarrow ln(1+ax+{bx}^2)-ln(1-ax+{bx}^2)$$$$\sim ax+{bx}^2-\frac{1}{2}{a}^2{x}^2-{abx}^3-\frac{b^2}{2}{x}^4+ax-{bx}^2+\frac{1}{2}{a}^2{x}^2-{abx}^3+\frac{1}{2}{b}^2{x}^4$$$$=2ax-2{abx}^3\Rightarrow f\left(x\right)\sim \frac{2ax-2{abx}^3}{\frac{x^2}{2}}(1-cosx\sim \frac{x^2}{2})\Rightarrow$$$$f\left(x\right)\sim \frac{4ax-4{abx}^3}{x^2}\Rightarrow f\left(x\right)\sim \frac{4a}{x}-4abx$$$$\Rightarrow {lim}_{x\to 0}f\left(x\right)=\mathrm{\infty }$$$$anotherwayln\left(\frac{1+ax+{bx}^2}{1-ax+{bx}^2}\right)=ln\left(\frac{1-ax+{bx}^2+2ax}{1-ax+{bx}^2}\right)=ln(1+\frac{2ax}{1-ax+{bx}^2})$$$$\sim \frac{2ax}{1-ax+{bx}^2}\Rightarrow f\left(x\right)\sim \frac{2ax}{(1-ax+{bx}^2)}\times \frac{1}{\frac{x^2}{2}}=\frac{4ax}{x^2(1-ax+{bx}^2)}=\frac{4a}{x(1-ax+{bx}^2)}$$$$\Rightarrow {lim}_{x\to 0}f\left(x\right)=\mathrm{\infty }$$

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