Calculating limits and asymptotes

Basic rules

Limits work well with operations: the limit of a sum is the sum of the limits, and likewise for the rest, provided the limits involved exist.

These rules also hold when $x\to \pm \infty$. The practical consequence is that, for polynomials, roots and quotients whose denominator is not zero, you just substitute: $\underset{x\to a}{\lim }P(x)=P(a)$ and $\underset{x\to a}{\lim }\frac{P(x)}{Q(x)}=\frac{P(a)}{Q(a)}$ if $Q(a)\ne 0$.

If the denominator gives $0$ and the numerator does not, there is no indeterminate form: the limit is $\pm \infty$ and is studied using the one-sided limits (there is a vertical asymptote).

Limits at a point: the indeterminate form $\frac{0}{0}$

If substituting $x=a$ gives zero for both numerator and denominator, the result $\frac{0}{0}$ tells you nothing: it is called an indeterminate form. It means that both have a common factor $(x-a)$, which must be removed.

Cancelling is legitimate when calculating a limit because, as $x$ approaches $a$, it never equals exactly $a$, and therefore $(x-a)\ne 0$.

Limits when $x\to \pm \infty$

With $x$ very large, in a polynomial the term of highest degree dominates: the others are negligible next to it. This gives a very short recipe.

For a quotient $\frac{P(x)}{Q(x)}$:

• if $\mathrm{degree}P<\mathrm{degree}Q$, the limit is $0$;
• if the degrees are equal, the limit is the quotient of the leading coefficients;
• if $\mathrm{degree}P>\mathrm{degree}Q$, the limit is $+\infty$ or $-\infty$ (decided by the sign of the quotient of the highest-degree terms).

For $x\to -\infty$, the sign depends on whether the degree is even or odd: $x^3\to -\infty$, but $x^4\to +\infty$.

When $\infty -\infty$ appears with roots, multiply and divide by the conjugate, as before, and then divide numerator and denominator by the highest power of $x$.

Asymptotes of a function

In a rational function, an oblique asymptote appears when the degree of the numerator exceeds that of the denominator by exactly one. If there is a horizontal asymptote on one side, there is no oblique asymptote on that side. For $n$ and $m$ there is a shortcut: divide the polynomials, and the quotient is $mx+n$.

Here $f(x)=\frac{x^2-x+1}{x-1}=x+\frac{1}{x-1}$: the term $\frac{1}{x-1}$ tends to $0$ far away from the origin, so the graph hugs the line $y=x$ (oblique asymptote) and shoots off near $x=1$ (vertical asymptote).