The limit rules as normally stated in introductory calculus texts are careful to avoid the problem of arithmetic involving infinity.  While there is no issue regarding limits at infinity, infinite limits present some concerns.  As a consequence, the problem $\underset{x\to \mathrm{\infty }}{lim}\left(x+1\right)$ cannot be solved using the sum rule as normally stated, as the sum rule requires that the component limits, $\underset{x\to \mathrm{\infty }}{lim}x$ and $\underset{x\to \mathrm{\infty }}{lim}1$ both exist, which means that they must be finite.  Since this clearly fails for $\underset{x\to \mathrm{\infty }}{lim}x$, some method other than the sum rule must be found to evaluate the original limit.

There are three approaches which could be taken to handle such situations: (1) Exclude such problems from the set which can be handled by the Student[Calculus1] package; (2) Introduce new rules to handle such cases; or (3) Extend the standard rules to cover cases where the necessary arithmetic involving infinities is unambiguous.  Maple takes the third approach.

Specifically, problems involving arithmetic combinations of terms, some of which have infinite limits at the limit point in question, can be divided into two groups: indeterminate forms and determinate forms.  Indeterminate forms are expressions such as $\underset{x\to \mathrm{\infty }}{lim}\frac{{\ⅇ}^x}{x}$ or $\underset{x\to 0+}{lim}{\left(x+1\right)}^{\frac{1}{x}}$, where care must be taken to understand which component dominates the overall computation.  Methods such as l'Hopital's Rule often must be used for these types of problems.  Determinate forms are arithmetic expressions involving infinity where either one term dominates (for example, only one component has an infinite limit) or where the effect of multiple infinities is to simply amplify each other.  (Note that it often happens that an indeterminate form can be transformed to a determinate form by algebraic manipulation.)

For example, $\underset{x\to \mathrm{\infty }}{lim}\left(x+{\ⅇ}^x\right)$ is a determinate form, because both terms are diverging to positive infinity, hence so is their sum.  Similarly, $\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}$ is a determinate form, since only one component, $\underset{x\to \mathrm{\infty }}{lim}x$ is diverging to infinity, while the other, $\underset{x\to \mathrm{\infty }}{lim}1$ remains bounded.

The limit rules used by Maple, when one or more component limits is infinity, are as follows.  For brevity and clarity, let $F=\underset{x\to a}{lim}f\left(x\right)$ and $G=\underset{x\to a}{lim}g\left(x\right)$, where at least one of $F$ or $G$ is infinite  Further, we will take "$F$ is finite" to mean "$\underset{x\to a}{lim}f\left(x\right)$ exists and is finite".  Then:

Constant multiple: For $c\ne 0$, $\underset{x\to a}{lim}cf\left(x\right)=cF$ for any $F$.

Sum: $\underset{x\to a}{lim}\left(f\left(x\right)+g\left(x\right)\right)=F+G$ if only one of $F$ or $G$ is infinite or if $F$ and $G$ are both infinite with the same sign.

Power: For $r\ne 0$, $\underset{x\to a}{lim}{f\left(x\right)}^r=F^r$ for any $F$.

Product: $\underset{x\to a}{lim}f\left(x\right)g\left(x\right)=FG$ if only one of $F$ or $G$ is infinite and the other is non-zero (or simply bounded away from 0 and of constant sign in a neighborhood of $x=a$) or both are infinite.

Quotient: $\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\frac{F}{G}$ if $F$ is infinite and $G$ is finite and non-zero (or simply bounded and of constant sign in a neighborhood of $x=a$) or $F$ is bounded and $G$ is infinite.

To complete the calculation of limits where any of the above rules has been applied and an infinite limit is involved, it is only necessary to be able to do some simple arithmetic involving infinity:

$\infty +\infty =\mathrm{\infty }$

$\infty \cdot \infty =\mathrm{\infty }$

$\frac{c}{\infty }=0$ for any constant $c$.

$c\cdot \infty =\mathrm{sign}\left(c\right)\cdot \mathrm{\infty }$ for any non-zero constant $c$, where $\mathrm{sign}\left(c\right)=\frac{c}{\left|c\right|}$ is the sign of $c$.

${\infty }^r=\mathrm{\infty }$ for any constant $0<r$.

${\infty }^r=0$ for any constant $r<0$.