indeterminate forms recall that, in some cases, attempting to evaluate a limit using substitution...
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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.
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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
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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.
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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).
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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
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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
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Trigonometric Limits
Let’s look at strategies for evaluating:
limx→0
tan xx lim
x→0
sin 4xx
lim xcos xx→0
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Techniques for Evaluating Limits
SubstitutionSimplification using factoringMultiplication by the conjugateAnalysis of infinite limitsSqueeze theoremChange of variables