\displaystyle \dfrac{1+2x+\dfrac{x^2}{2}+...}{-1-\frac{x^2}{2}+\cdots}[br]\[br]= (-1)(1+2x+\dfrac{x^2}{2} + \cdots)(1+ (\dfrac{x^2}{2} + \dfrac{x^3}{6}+\cdots))^{-1}
\displaystyle F(x) = \dfrac{e^x}{1-x}[br]\[br]\text{Show that} F(x) \to -\infty \ \text {as} \ x \to \infty
\displaystyle \dfrac{1+2x+\dfrac{x^2}{2}+...}{-1-\frac{x^2}{2}+\cdots}[br]\[br]= (-1)(1+2x+\dfrac{x^2}{2} + \cdots)(1+ (\dfrac{x^2}{2} + \dfrac{x^3}{6}+\cdots))^{-1}
\displaystyle F(x) = \dfrac{e^x}{1-x}[br]\[br]\text{Show that} F(x) \to -\infty \ \text {as} \ x \to \infty
\displaystyle \dfrac{1+2x+\dfrac{x^2}{2}+...}{-1-\frac{x^2}{2}+\cdots}[br]\[br]= (-1)(1+2x+\dfrac{x^2}{2} + \cdots)(1+ (\dfrac{x^2}{2} + \dfrac{x^3}{6}+\cdots))^{-1}
\displaystyle F(x) = \dfrac{e^x}{1-x}[br]\[br]\text{Show that} F(x) \to -\infty \ \text {as} \ x \to \infty
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