Answer
$-\dfrac{1}{2}$
Work Step by Step
Given: $\lim\limits_{x\to\infty}\frac{x+x^2}{1-2x^2}$
Here, $x\to\infty$, $x+x^2$ approaches $\infty$ and $1-2x^2$ approaches $-\infty$.
This shows an indeterminate form of type $\infty/-\infty$, and so we will apply L'Hospital Rule.
$\lim\limits_{x\to\infty}\dfrac{\frac{d}{dx}(x+x^2)}{\dfrac{d}{dx}(1-2x^2)}=\lim\limits_{x\to\infty}\dfrac{1+2x}{-4x}$
On dividing both numerator and denominator by $x$, we get the highest power in the denominator, so
$\lim\limits_{x\to\infty}\dfrac{\frac{1+2x}{x}}{\dfrac{-4x}{x}}=\lim\limits_{x\to\infty}\dfrac{\frac{1}{x}+2}{-4}$
As we know $\lim_{x\to\infty}(\frac{1}{x})=0$.
Thus, we have
$=\dfrac{0+2}{-4}$
or, $=-\dfrac{1}{2}$
