Limit of rational function (ratio of 2 polynomials, p(x) & q(x))
limx→c( p(x) / q(x) ), given q(c) ≠ 0
= p(c) / q(c)
Because
Quotient rule
Limit of polynomial
Sandwich Theorem?
Examples
limx→0( x cosx )
(at least) 2 ways of figuring this out, the first of which demonstrates
Sandwich Theorem
-|x| ≤ x cosx ≤ |x| because -1 ≤ cos x ≤ 1
limx→0-|x| = limx→0|x| = 0
So Sandwich Theorem says limx→0( x cosx ) = 0
Or
limx→0( x cosx )
= limx→0(x) limx→0(cosx )
= 0 × 1 = 0
Or just plug x = 0 into x cosx
You can do this because x cosx is continuous around x = 0,
i.e., has no breaks in its graph
If f(x) is continuous around c then limx→cf(x) = f(c)
But technically using continuity like this to find a limit
is circular, because the definition of continuity is that
the limit equals the function value—but as long as we’re
doing limits intuitively, continuity is an easy property to
recognize by intuition.
If limx→0f(x) = 3 and limx→0g(x) = 16, what is
limx→0( f(x) + g(x) )?
Sum rule: 3 + 16 = 19
limx→0( √g(x) )?
root rule: √16 = 4
Definition of Limit
Think of idea of limit as where function seems to be heading across a small gap in its graph
If function heads for the same value no matter how small you make the gap, then that value is the limit
limx→cf(x) = L iff for any ε > 0 you can
find a δ such that whenever 0 < | x - c | < δ,
then 0 < | f(x)- L | < ε