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# Limits of Trigonometric Functions

#### theorem

Theorem :
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 ,
then
lim_(x ->a) f(x) ≤ lim_(x -> a) g(x).
The explain in following fig.

#### theorem

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.,1/2OA.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

Theorem - The following are two important limits.
i) lim_(x -> 0) sin x / x = 1

ii) lim_(x->0) (1 - cos x ) / x = 0

Proof :
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)
Then

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

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Sandwich Theorem [00:06:46]
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