Define the function $f$

$f\left(x\right)\=\frac{3x^5\+1}{{\left|x\right|}^5\+5x^2\+1}\+\frac{10\sin \left(x\right)}{x^2\+1}$$\stackrel{\text{assign as function}}{\to }$

Graphical analysis

Application of Maple's limit operator

$\underset{x\to \infty }{lim}f\left(x\right)$ = $\underset{x\→\mathrm{\∞}}{lim}f\left(x\right)$$$

$\underset{x\to -\infty }{lim}f\left(x\right)$ = $\underset{x\→-\mathrm{\∞}}{lim}f\left(x\right)$$$

Solution from first principles

Application of the Sum rule is valid if the limit of each summand exists. There are two summands:

$g\left(x\right)\=\frac{3x^5\+1}{{\left|x\right|}^5\+5x^2\+1}$ and $h\left(x\right)\=\frac{10\sin \left(x\right)}{x^2\+1}$ 

The limits at $\pm \infty$ for $h\left(x\right)$ are handled by Principle 1.1.1 or the Squeeze theorem, as per Example 1.5.4. These limits will certainly be zero.

In evaluating the limit at $\+\infty$ for $g\left(x\right)$, the absolute value in the denominator can be ignored because $x\>0$ for large $x$. Hence, the limit is that of a rational function in which both numerator and denominator have the same degree. The limit will be the ratio of the leading coefficients of the numerator and denominator, that is, 3.

In evaluating the limit at $-\infty$ for $g\left(x\right)$, the absolute value in the denominator is replaced by $-x$, so that again, the limit is taken of a rational function, namely, the rational function

 $\frac{3x^5\+1}{-x^5\+5x^2\+1}$ 

Dividing the numerator and denominator of this fraction by $x^5$, the degree of the denominator, leads to

$\frac{3\+\frac{1}{x^5}}{-1\+\frac{5}{x^3}\+\frac{1}{x^5}}$ 

from which the limit $-3$ is readily obtained.