Limit Theorems

Definition If $p$ is a polynomial, then $\lim_{x \to a} p(x) = p(a)$ for every real $a$. If $r = p/q$ with $p$ and $q$ polynomials and $q(a) \neq 0$, then

Both statements follow from the limit laws together with $\lim_{x \to a} c = c$ and $\lim_{x \to a} x = a$.

Definition If $f(x) = g(x)$ for every $x \neq a$ in some open interval containing $a$, and $\lim_{x \to a} g(x)$ exists, then $\lim_{x \to a} f(x)$ exists and equals it. For a rational function with $p(a) = q(a) = 0$, the factor $x - a$ divides both $p$ and $q$, and cancelling it gives such a $g$.

Definition Rationalization multiplies the numerator and the denominator of a quotient by the conjugate $\sqrt{A} + \sqrt{B}$, which is allowed wherever $\sqrt{A} + \sqrt{B} \neq 0$, and then uses the identity

The result is an expression equal to the original for $x \neq a$ near $a$, often one where a factor can be cancelled.

Definition If $f(x) \le g(x)$ for every $x \neq a$ in some open interval containing $a$, and both $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist, then

A strict inequality $f(x) < g(x)$ near $a$ still only gives $\le$ for the limits. A related fact is sign preservation. If $\lim_{x \to a} f(x) = L > 0$, then $f(x) > 0$ for every $x \neq a$ in some open interval around $a$.

Definition Suppose $g(x) \le f(x) \le h(x)$ for every $x \neq a$ in some open interval containing $a$, and

Then $\lim_{x \to a} f(x) = L$. Note that $f$ is not assumed to have a limit. The theorem proves that it does.

Definition A function $g$ is bounded near $a$ if there are numbers $\delta > 0$ and $K$ such that $\lvert g(x) \rvert \le K$ whenever $0 < \lvert x - a \rvert < \delta$. If $\lim_{x \to a} f(x) = L$ exists as a real number, then $f$ is bounded near $a$. Indeed, the definition of the limit with $\varepsilon = 1$ gives $\delta > 0$ such that $0 < \lvert x - a \rvert < \delta$ implies

The converse is false. A consequence used often is this. If $g$ is bounded near $a$ and $\lim_{x \to a} f(x) = 0$, then $\lim_{x \to a} f(x) g(x) = 0$. The reason is that $0 \le \lvert f(x) g(x) \rvert \le K \lvert f(x) \rvert$ near $a$, and $K \lvert f(x) \rvert \to 0$, so the squeeze theorem gives $\lvert f(x) g(x) \rvert \to 0$, which is the same as $f(x) g(x) \to 0$.