Limits

Stage 18 of 23 Strand 1 of 2 8 lessons

8 illustrated lessons, each teaching the why before the how.

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The Idea of a Limit

What the function closes in on, arrival or not.

A limit is the value a function approaches, whether or not it ever gets there

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00.511.52

Halve the gap each time. The values approach 2 without ever reaching it.

xy

A limit at x = 2 asks what happens near 2, never what happens exactly at 2.

23.93.994.14.014, with no value at 2

This function is undefined at x = 2, yet it approaches 4. The hole does not stop the limit.

Now you

00.511.52

What does 2 + 1/n as n grows without bound approach?

00.511.52

What does 1/2 + 1/4 + 1/8 + … continuing forever approach?

One Sided Limits

The left and the right may disagree.

Approaching from the left and from the right can give two different answers

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xy

Approaching from the left, the values settle on −1. From the right, they settle on +1.

−2−1012left −1right +1

The two sides give different answers, so there is no single value to approach.

limit existsand it fails the moment they disagreeboth sides agree

A limit only exists when the two sides agree. That is the whole condition.

Now you

xy

The left limit is 0 and the right limit is 0. Does the limit exist?

xy

The left limit is 3 and the right limit is 3. Does the limit exist?

The Limit Laws

Limits pass through sums, products and quotients.

Limits pass straight through sums, products and quotients

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lim (f + g)each piece may be taken separatelylim f + lim g

The limit of a sum is the sum of the limits, so you may take each limit separately.

lim (f × g)the same rule, for productslim f × lim g

Products behave the same way: the limit of a product is the product of the limits.

only when the limit of g is not 0

Quotients carry one condition: the limit of the denominator must not be zero.

Now you

is the law allowed here?

f → 8 and g → 0. May the quotient law be used on f / g?

lim(f × g)what does this come to?

If f → 3 and g → 6, what does f × g approach?

Indeterminate Forms

Zero over zero decides nothing on its own.

Zero over zero decides nothing, so the expression has to be rewritten first

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23.93.994.14.01(x² − 4) ÷ (x − 2) at x = 2 is 0 ÷ 0

Substitute x = 2 into (x² − 4) / (x − 2) and you get 0 over 0 — a signal, not an answer.

factor the numerator
cancel — valid everywhere except x = 2
x + 2

Factor the numerator and the (x − 2) cancels, because x is near 2 and never equal to 2.

23.93.994.14.01now the value at 2 is 4 as well

Now substitute x = 2 into x + 2. The limit is 4.

Now you

what does this come to?

What is the limit of (x² − 9) / (x − 3) as x → 3?

what does this come to?

What is the limit of (x² − 16) / (x − 4) as x → 4?

Limits at Infinity

Far out, only the highest power matters.

For very large x only the highest power matters, so the rest can be ignored

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xy

This curve climbs toward 2 and then flattens off, never quite reaching it.

divide numerator and denominator by x
the 1/x terms vanish as x grows
2 ÷ 1 = 2

Divide numerator and denominator by the highest power of x present.

∞1.911.99heading for 2

Every 1/x term shrinks to 0, leaving the ratio of the leading coefficients.

Now you

what does this come to?

What is the limit of (2x + 3) / (3x + 5) as x → ∞?

what does this come to?

What is the limit of (9x + 3) / (8x + 5) as x → ∞?

The Squeeze Theorem

Trapped between two functions sharing a limit.

A function trapped between two others that share a limit must share that limit

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xy

Suppose one function sits above another, with a third caught between them.

xy

If the upper and lower functions approach the same value, the middle one has nowhere else to go.

xy

That is how a function that oscillates endlessly near 0 can still have a single limit there.

Now you

y4

f is trapped between two functions that both approach 4. What does f approach?

y2

f is trapped between two functions that both approach 2. What does f approach?

The Limit of sin x over x

The ratio that unlocks the trig derivatives.

Near 0, sin x over x approaches 1 — the limit the trigonometric derivatives rest on

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at x = 0
not an answer: the limit must be found another way
0 ÷ 0

Substituting x = 0 gives 0 over 0, which is not an answer: the limit must be found another way.

xy

As x approaches 0 the curve approaches height 1, but the point at x = 0 is missing.

near 0
and
the squeeze — the middle has nowhere else to go

Near 0 the ratio is trapped between cos x and 1, and both of those approach 1.

in radians — this is why calculus uses radians
near 0

In radians sin x is very close to x near 0, which is what makes d/dx sin x = cos x work.

Now you

xy

The squeeze proof traps sin x / x between which two functions?

xy

Why must x be in radians?

Small-Angle Approximations

Near zero, the curves flatten into polynomials.

Near 0, sine, cosine and tangent can each be replaced by a simple polynomial

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xy

Near 0, sin x is close to x. Drawn together, the curve and the line almost coincide.

xy

Cosine falls away from 1 like the parabola 1 − x²/2: flat at its maximum, then falling.

the numerator is near x, the denominator is near 1

Tangent follows: a numerator near x over a denominator near 1 is nearly x itself.

sin 0.1compare it with the approximation 0.10.09983…

At x = 0.1 the approximation 0.1 matches sin x to three decimal places.

Now you

xy

For small x, cos x ≈ ?

xy

The approximations hold only when x is measured in which unit?

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