Let f and g be two real valued functions with the same domain such that f(x) ≤ g(x) for all x in the domain of definition, For some a, if both
`lim_( x -> a)` f(x) and `lim _(x - >a)` g(x) exist ,
`lim_(x ->a)` f(x) ≤ `lim_(x -> a)` g(x).
The explain in following fig.
Theorem : (Sandwich Theorem)
Let f, g and h be real functions such that f (x) ≤ g( x) ≤ h(x) for all x in the common domain of definition. For some real number a , if
`lim_(x - > a)`f(x) = l = `lim_(x ->a)` h(x) , then `lim_(x ->a)` g(x) = l .fig.
Given below is a beautiful geometric proof of the following important inequality relating trigonometric functions.
`cos x < sin x/x < 1 for 0 <|x| < pi / 2`
Proof : We know that sin (– x) = – sin x and cos( – x) = cos x. Hence, it is sufficient to prove the inequality for 0 < x < `pi /2`. In the following fig.
O is the centre of the unit circle such that the angle AOC is x radians and 0 < x <`pi /2` .Line segments B A and CD are perpendiculars to OA. Further, join AC. Then
Area of OAC ∆ < Area of sector OAC < Area of ∆ OAB .
i.e.,`1/2`OA.CD <`x / 2pi` .`( OA) ^2` < `1/2` OA .AB.
i.e., CD < x . OA < AB.
From ∆ OCD,
sin x = `(CD)/(OA)` (since OC = OA) and hence CD = OA sin x. Also tan x = `(AB)/(OA)` and hence AB = OA. tan x.
Thus OA sin x < OA. x < OA. tan x.
Since length OA is positive,
we have sin x < x < tan x.
Since 0 < x < `pi/2` , sinx is positive and thus by dividing throughout by sin x, we have `1 < x/(sin x) < 1/(cos x)` . Taking reciprocals throughout , we have
`cos x < (sin x)/x < 1`
which complete the proof.
Theorem - The following are two important limits.
i) `lim_(x -> 0) sin x / x = 1`
ii) `lim_(x->0) (1 - cos x ) / x = 0`
i) The inequality in (*) says that the function `sin x / x ` is sandwiched between the functions cos x and the constant function which takes value 1 .
Further, since `lim _(x→0)` cos x = 1, we see that the proof of (i) of the theorem is complete by sandwich theorem.
ii) we recall the trigonometric identity
1 – cos x = 2 `sin^2 (x / 2)`
`lim_(x -> 0) (1 - cos x)/x = lim_(x -> 0) 2 sin^2 (x/2) / x = lim_(x-> 0) sin (x / 2) / (x / 2) . sin (x / 2)`
`lim_(x -> 0) sin(x / 2) / (x / 2) . lim_(x->0) sin (x/2) = 1.0 = 0`
Observe that we have implicitly used the fact that 0 x → is equivalent to `x/2 -> 0` . This may be justified by putting ` y = x / 2`