Limits at Negative Infinity

So far we have discussed limits at some fixed numbers. In this section we shall be concerned with the limits at negative infinity.

Let us consider the function $f$ defined by the equation $f\left( x \right) = \frac{{2{x^2}}}{{{x^2} + 1}}$. Let $x$ take on the values 0, -1, -2, -3, -4, -5, -10, -100, -1000 etc., allowing $x$ to decrease through negative values without bound. In this case, we say that $x$ is approaching negative infinity. We write it as $x \to – \infty$, as shown in the table, and it gives the corresponding function values, $f\left( x \right)$. The corresponding function values are given in the table below.

 $x$ $0$ $– 1$ $– 2$ $– 3$ $– 4$ $– 5$ $– 10$ $– 100$ $– 1000$ $f\left( x \right) = \frac{{2{x^2}}}{{{x^2} + 1}}$ $0$ $1$ $\frac{8}{5}$ $\frac{9}{5}$ $\frac{{32}}{{17}}$ $\frac{{25}}{{13}}$ $\frac{{200}}{{101}}$ $\frac{{20000}}{{10001}}$ $\frac{{2000000}}{{1000001}}$

We note that the function values are the same for the negative numbers as for the corresponding positive numbers. So, we can see that $f\left( x \right) \to 2$ as $x \to – \infty$, i.e.
$\mathop {\lim }\limits_{x \to – \infty } \frac{{2{x^2}}}{{{x^2} + 1}} = 2$