Indeterminate Forms Recall that, in some cases, attempting to evaluate a limit using substitution yields an indeterminate form, such as 0/0. Usually, we can use factoring if the function is a rational function… … or we can multiply by the conjugate if the function has radicals, etc.

sin x lim x Indeterminate Forms But what if the function and the limit involve trigonometry, such as: Substitution yields an indeterminate form, but can we rewrite our expression to “fix” the 0/0 issue? lim x→0 sin x x

Indeterminate Forms Graphically, this limit clearly exists (and equals 1). We cannot use any of our previously studied methods to “fix” the function. Let’s look at other functions with similar behavior around x=0.

The Squeeze Theorem In the neighborhood of x=1, the graph of f(x)=(sin x)/x is “squeezed” by the graphs of g(x)=cos x and h(x)=1. Because the limits of g(x) and h(x) both equal 1 as x approaches 0, so must that limit on f(x).

lim g(x) = lim h(x) = L lim f(x) = L The Squeeze Theorem Let f, g, and h be defined on an interval containing c (except possibly at c itself). For every x other than c in that interval, g(x)<f(x)<h(x). If , then lim g(x) = x→c lim h(x) = L x→c lim f(x) = L x→c

lim = 1 sin x x lim = 0 1 – cos x x Trigonometric Limits The squeeze theorem is useful for finding trigonometric limits, including: But it’s not always necessary… lim = 1 x→0 sin x x lim = 0 x→0 1 – cos x x

tan x sin 4x lim lim x x lim xcos x Trigonometric Limits Let’s look at strategies for evaluating: lim x→0 tan x x lim x→0 sin 4x x lim xcos x x→0

Techniques for Evaluating Limits Substitution Simplification using factoring Multiplication by the conjugate Analysis of infinite limits Squeeze theorem Change of variables