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**HallsofIvy** You could also use the MacLaurin series for $\displaystyle e^x$, $\displaystyle 1+ x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot$ so that $\displaystyle 1- e^x= -(x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot}$.

Now you have $\displaystyle \frac{x^2}{1- e^x}= -\frac{x^2}{x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot}$.

Divide both numerator and denominator by $\displaystyle x^2$ to get $\displaystyle -\frac{1}{\frac{1}{x}+ \frac{1}{2}+ \frac{x}{6}+ \cdot\cdot\cdot}$ which goes to infinity as x goes to infinity. Notice that, by this argument, $\displaystyle \lim_{x\to\infty} \frac{x}{1- e^x}= 1$.