Mathematical Proof: Limits of Trig Functions - sin(x)/x & Squeeze Theorem, Slides of Pre-Calculus

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The remarkable limit

lim

x→ 0

sin(x)

x

Trigonometric functions revisited

Trigonometric functions like sin(x) and cos(x) are continuous everywhere. Informally, this can be explained as follows: a small perturbation of a point on the unit circle results in small changes in its x- and y -coordinates.

x

y

These functions are periodic, and so have an oscillating behaviour at infinity. Therefore, they have neither a finite nor an infinite limit at infinity.

The Squeeze Theorem

Interesting things start to happen when me mix trigonometric and polynomial functions. For instance, one of the most important limit for applications of calculus is lim x→ 0

sin x x. So far we have not proved any results that would allow to approach this limit. Theorem. lim x→ 0

sin x x = 1. Informal proof. The key idea of the proof is very simple but very important. Suppose that we have three functions f (x), g (x), and h(x), and that we can prove that: (^1) the inequalities g (x) ≤ f (x) ≤ h(x) hold for all x in some open interval containing the number c, possibly excluding c itself; (^2) lim x→c g (x) = lim x→c h(x) = L. Then lim x→c f (x) = L as well. This is The Squeeze Theorem: the values of f are “squeezed” between values of g and h. It is also called The Sandwich Theorem, or, in some languages, The Two Policemen (and a Drunk) Theorem.

The Squeeze Theorem: illustration

Proving lim

x→ 0

sin x

x = 1

Why sin x ≤ x ≤ tan x? This is proved using the geometric picture

x

y

where we can actually find all the quantities involved!

Proving lim

x→ 0

sin x

x = 1

Indeed,

x

y

x

1

the area of the small triangle is 12 · 1 · 1 · sin x = 12 sin x;

Proving lim

x→ 0

sin x

x = 1

x

y

x

1

tan x

and the area of the large triangle is 12 · 1 · tan x = 12 tan x.

Proving lim

x→ 0

sin x

x = 1

Therefore an obvious inequality between the areas

x

y

implies sin x ≤ x ≤ tan x, which is what we needed.

Consequences of lim

x→ 0

sin x

x = 1

lim x→ 0 +

sin x x^2 =^ xlim→ 0 +

[sin x x ·^

1 x

]

lim x→ 0

tan x x = lim x→ 0

[ (^) sin x x ·^

1 cos x

]

lim x→ 0

sin 2x x = lim x→ 0 2

sin 2x 2 x =^2 xlim=t→ 0 2

sin t t = 2.

lim x→ 0

1 −cos x x^2 = lim x→ 0

1 −(cos x)^2 (1+cos x)x^2 = lim x→ 0

(sin x)^2 (1+cos x)x^2 = lim x→ 0

[(

sin x x

1+cos x

]

so lim x→ 0

1 −cos x x^2 = lim x→ 0

( (^) sin x x

lim x→ 0

1 1+cos x = 1

2 =^

1

lim x→ 0

1 −cos x x = lim x→ 0

[ (^1) −cos x x^2 ·^ x

]

lim x→ 0

2 −cos 3x−cos 4x x = lim x→ 0

[

3 1 −cos 3 3 x x+ 4 1 −cos 4 4 x x

]

Informal consequences of lim

x→ 0

sin x

x = 1

(^1) lim x→ 0

sin x x = 1 means informally that for small^ x^ we have sin^ x^ ≈^ x, (^2) lim x→ 0

tan x x = 1 means informally that for small^ x^ we also have tan x ≈ x, (^3) lim x→ 0

1 −cos x x^2 =^

1 2 means informally that for small^ x^ we have cos x ≈ 1 − x 2

These approximate formulas give examples of a general strategy of differential calculus: replacing a function by a polynomial expression that approximates it very well for small x (or x close to the given point a).