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.

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?

limx→0

sin xx

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).

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) = Lx→c

lim f(x) = L x→c

Trigonometric Limits

The squeeze theorem is useful for finding trigonometric limits, including:

But it’s not always necessary…

lim = 1x→0

sin xx lim = 0

x→0

1 – cos xx

Trigonometric Limits

Let’s look at strategies for evaluating:

limx→0

tan xx lim

x→0

sin 4xx

lim xcos xx→0

Techniques for Evaluating Limits

SubstitutionSimplification using factoringMultiplication by the conjugateAnalysis of infinite limitsSqueeze theoremChange of variables