Short and Sweet Calculus

## 2.5 Theorems for Calculating Limits

In this section, we discuss how we can evaluate limits analytically. We can evaluate many limits by applying the following limit laws.

2.2. (Algebraic Operations on Limits): Suppose $\lim_{x\to s}f(x)=L\qquad\text{and}\qquad\lim_{x\to s}g(x)=M,$ where $$L$$ and $$M$$ are two real numbers, and $$s$$ signifies $$a,a^{+},a^{-},-\infty$$, or $$+\infty$$ with $$a$$ being a real number. Then

1. ($$k$$ is a constant number)

2. $$\lim_{x\to s}\left[f(x)+g(x)\right]=\lim_{x\to s}f(x)+\lim_{x\to s}g(x)}=L+$$

3. $$\lim_{x\to s}\left[f(x)-g(x)\right]=\lim_{x\to s}f(x)-\lim_{x\to s}g(x)=L-M$$

4. $$\lim_{x\to s}\left[f(x)g(x)\right]=\left(\lim_{x\to s}f(x)\right)\left(\lim_{x\to s}g(x)\right)=LM$$

5. $$\lim_{x\to s}\frac{f(x)}{g(x)}=\frac{\lim_{x\to s}f(x)}{\lim_{x\to s}g(x)}=\frac{L}{M}$$ (provided $$M\neq0$$)

This theorem merely says:

1. The limit of a constant times a function is the constant times the limit of the function.

2. The limit of a sum is the sum of the limits.

3. The limit of a difference is the difference of the limits.

4. The limit of a product is the product of the limits.

5. The limit of a quotient is the quotient of the limits provided that the limit of the denominator is not zero.

2.3. (Limits of Polynomials and Rational Functions): Let $$P(x)$$ and $$Q(x)$$ be two polynomials and $$a$$ be a real number. Then
1. Polynomial functions: $\lim_{x\to a}P(x)=P(a)$ 2. Rational functions: $\lim_{x\to a}\frac{P(x)}{Q(x)}=\frac{P(a)}{Q(a)}\quad(\text{provided }Q(a)\neq0)$ Similar results hold for left and right hand limits; that is, we can replace $$x\to a$$ by $$x\to a^{-}$$ or $$x\to a^{+}$$ in the above equations.

2.4. Let $$a$$ and $$L$$ be real numbers. If $${\lim_{x\to a}g(x)=0}$$ and $${\lim_{x\to a}f(x)=L\neq0}$$ then $\lim_{x\to a}\frac{f(x)}{g(x)}=\begin{cases} +\infty & \text{if }L>0\text{ and }g(x)>0\\ -\infty & \text{if }L>0\text{ and }g(x)<0\\ -\infty & \text{if }L<0\text{ and }g(x)>0\\ +\infty & \text{if }L<0\text{ and }g(x)<0 \end{cases}$ The theorem is also valid for the left and right-hand limits or limits at $$+\infty$$ or $$-\infty$$; that is, if we replace $$x\to a$$ by $$x\to a^{-}$$, $$x\to a^{+}$$, $$x\to+\infty$$, or $$x\to-\infty$$.

• In the above theorem, we talk about the sign of $$g(x)$$ for all $$x$$ close to $$a$$ (not in its entire domain); it does not matter if the sign of $$g$$ changes when $$x$$ is NOT very close to $$a$$.

We may summarize the above theorem and symbolically write

$\frac{L}{0}=+\infty\text{ or }-\infty\qquad(\text{if }L\neq0)$ The result depends on the sign of $$L$$ and the sign of $$g(x)$$ for $$x$$ close to $$a$$. Recall that $\frac{(+)}{(+)}=\frac{(-)}{(-)}=(+)\quad\text{and}\quad\frac{(-)}{(+)}=\frac{(+)}{(-)}=(-).$

• Notice that $$L/0=\pm\infty$$ is just a symbol for memorizing Theorem 2.4. We do not divide $$L$$ by zero (division by zero is not defined) because the denominator is not exactly zero! The denominator gets closer and closer to $$0$$ as $$x$$ gets closer and closer to $$a$$, but not equal to $$a$$. If $$g(x)=0$$ for all $$x$$ close to $$a$$, then the function $$h(x)=f(x)/g(x)$$ will not be defined near $$a$$ and so we cannot talk about the limit of $$f(x)/g(x)$$ when $$x$$ approaches $$a$$.