# Evaluate the limit of the function f=(1-cosx)/x^2 x-->0.

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f= (1-cosx)/x^2 x --> 0

Let us substitute:

lim f(x) = 1-1/0 = 0/0 ( then the method has failed)

But we know that:

We know that cos2x = cos^2 x - sin^2 x

==> cosx = cos^2 (x/2) - sin^2 (x/2)

==+ -cosx = sin^2 (x/2) - cos^2 (x/2)

==> 1 - cosx = sin^2 (x/2) - cos^2 (x/2) + 1

But cos^2 (x/2) = 1- sin^2 (x/2)

==> 1- cosx = sin^2 (x/2) -1 + sin^2 (x/2) + 1

==> 1-cosx = 2sin^2 (x/2)

Now let us substitute:

lim f(x) = lim(2sin^2 (x/2))/x^2

= 2lim(sin^2 (x/2) /x^2

Let us multiply and divide by (x/2)^2

==> lim f(x) = 2 lim (sin^2 (x/2)/(x/2)^2]*(x/2)^2/ x^2

= 2lim x^2/4[(sin (x/2) / (x/2)]^2 / x^2

= (1/2) lim sin[(x/2) /(x/2)]

But we know that lim sinx/x = 1

==> lim f(x) = (1/2) * 1= 1/2

==> **lim f(x) = 1**

To find the limit of f(x) = (1-cosx)/x^2 as x --> 0

Solution:

We can prove this by L Hospitals rule as lt (1-cosx)/x^2 has a 0/0 inditerminate form.

L hospital's rule: Differetiate numerator and differentiate denominator and then take limits.

Ltf(x) = lt (1-cosx)/x^2 is 0/0 form as x--> 0.

lt f(x) = lt (1-cosx)'/(x^2)' = lt (sinx /2x) =(1/2) Lt (sinx)/x = 1/2,

as lt(sinx)/x = 1 is a standard limit.

2nd method:

We know that lt (sinx )/x asx-->0 = 1..........(1)

Cos(a+b) = cosacosb - sinasinb.

Cosx = cos(x/2+x/2) = Cos^2(x/2)-sin^2(x/2)

cosx = 1-cos^2(2/2)-cos^2(x/2)

cosx = 1-2sin^2 (x/2)

Therefore (1-cosx)/x^2 = (1- (1-2sin^2(x/2)}/x^2 = (2sinx^2)/x^2

Taking limits as x--> 0, we get:

lt(1-cosx)/x^2 = lt2sin^2x/x^2

lt(1-cosx) = 2 lt [(sinx/2)/x]{(sinx/2)/x}

lt (1-cosx) = 2lt [(sint)/2t][sint/2t] as t-->0, where t= x/2

lt(1-cosx) = (2/4)[lt[(sint)/] [lt(sint)/t]

lt (1-cosx) = (2/4)[1][1] by virtue of the limit at (1).

Therefore lt (1-cosx) = 1/2.

First, we'll verify if it is a case of indeterminacy, so we'll substitute x by 0, into the given expression:

f(0)=(1-cos0)/0^2, where cos0 = 1

f(0) = (1-1)/0^2

f(0) = 0/0, indeterminacy case

We'll solve the limit, by substituting the difference

1-cos x = 2 [sin(x/2)]^2

lim (1-cos x)/x^2 = lim 2 [sin(x/2)]^2 / x^2

2*lim [sin(x/2)]^2 / x^2 = lim [sin (x/2)/x]*lim [sin (x/2)/x]

We know that lim (sin x)/x = 1

lim [sin (x/2)/2*(x/2)] = (1/2)lim [sin (x/2)/(x/2)] = 1/2

lim (1-cos x)/x^2 = 2*(1/2)*1*(1/2)*1 = 1/2

**lim (1-cos x)/x^2 = 1/2 , when x->0**

Another method of solving the limit is to apply l'Hospital rule, because, after evaluation, we've obtained an indetermacy case "0/0".

lim (1-cos x)/x^2 = lim (1-cos x)'/(x^2)'

where

(1-cos x)' = 0 - (-sin x) = sin x

(x^2)' = 2x

lim (1-cos x)/x^2 = lim (sin x) / 2x = 1/2 * lim (sinx)/x = 1/2 * 1

**lim (1-cos x)/x^2 =1/2**