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Limits of Trigonometric Functions

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Limits of Trigonometric Functions - Lesson Summary

Evaluating Trigonometric Functions

Theorem: Let g and h be two real valued functions with the same domain, such that g(x) ≤ h(x) for all x in the domain of definition.

For some a, if both   lim x → a g(x)   and    lim x → a h(x)   exist, then   lim x → a g(x) ≤   lim x → a h(x)

The result of this theorem is used for another theorem, called the sandwich theorem.

The sandwich theorem is highly useful in finding the limits of trigonometric functions otherwise tedious to analyse.


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

In the figure, the curves in black represent the functions, 'f(x)' and 'h(x)'.

The red curve represents the function, 'g(x)'.

At point 'a,' the limit of the functions, 'f of x', 'g of x' and 'h of x,' is supposed to be 'l'.



Limits of Trigonometric Functions

Consider a unit circle.

OQ is a line equal to the length of the radius. Draw another line OS making an angle x radians with line OQ.

x should lie between zero and π/2. Draw the arc.

Draw a perpendicular from point S on to line OQ. The meeting point is P.

Next, draw a line from Q and perpendicular to line OQ.

Extend line OS to meet this perpendicular line at R.

Area of ΔOQS < Area of sector ΔOQS < Area of ΔOQR

½ . OQ . PS < x/2π .π. OQ2 < ½ . OQ . QR

All the quantities are real numbers.

Therefore, PS < x .OQ < QR.

In ΔOSP, sin x = PS/OS

sin x = PS/OQ [since, OS = OQ = Radius]

⇒ PS = OQ.sin x

In ΔOQR, tan x = QR/OQ

PS < x.OQ < QR


Hence, QR = OQ. tan x

OQ.sin x < x.OQ < OQ.tan x

Since OQ is positive, sin x < x > tan x.

Since 0 < x < π/2, sin x is positive.

1 < x/sin x <1/cos x

cos x < sin x/x < 1

Here, the function (sin x) / x is sandwiched between cos x and 1.

  lim x → 0  sin x x   = 1  
  lim x → 0  1 - cos x x  

lim x → 0  sin x x   = 1

cos x < sin x /x < 1

  lim x → 0 Cos x  <   lim x → 0  sin x x   < lim x → 0  1

At zero, the limit of the function, cos x, is one.

lim x → 0 Cos x =1

The limit of the constant, one, is one.

lim x → 0 1 = 1

Therefore, 1  <  lim x → 0  sin x x    < 1

Therefore, according to the sandwich the theorem, the limit of the function, sin x divided by x, is sandwiched.

lim x → 0  sin x x   =1

lim x → 0  1 - cos x x  


  lim x → 0  2 sin2 (x/2) x                       [Since1 - cos x = 2 sin2 (x/2)]

= lim x → 0  sin2 (x/2) x/2


= lim x → 0   sin (x/2) x/2   x sin (x/2)  

= lim x → 0   sin (x/2) x/2   x  lim x → 0  sin (x/2)

['.'  lim x → a [f(x).g(x)] = lim x → a f(x) . lim x → a g(x) ]

As x → 0 , we have x/2 → 0

= lim x/2 → 0   sin (x/2) x/2   x  lim x/2 → 0  sin (x/2)

= 1 x  lim x/2 → 0  sin (x/2)    [By the sandwich theorem, the limit in the first function is equal to one.]

= 1 X 0 = 0  [The limit of the second function is obviously zero.}

Therefore, the second proposition is proved.

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