I need help to solve using remarcable limits.limit of f(x) given by f(x)=(1-cosx)/x^2*cos4x, x->0

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll re-write the numerator of the function f(x) using the half angle identity.

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

f(x) = 2[sin(x/2)]^2/x^2*cos4x

We'll evaluate the limit:

lim 2[sin(x/2)]^2/x^2*cos4x = 2lim[sin(x/2)]^2/x^2*lim 1/cos4x

2lim[sin(x/2)]^2/x^2*lim 1/cos4x = 2lim[sin(x/2)]/x*lim sin(x/2)/x*lim 1/cos4x

We know that the remarcable limit is:

lim sin u(x)/u(x) = 1

We'll create the remarcable limit:

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

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

The limit of the function is:

2lim[sin(x/2)]^2/x^2*lim 1/cos4x = 2*(1/2)*(1/2)*lim 1/cos4x

We'll substitute x by 0:

2*(1/2)*(1/2)*lim 1/cos4x = (1/2)*(1/cos 0) = 1/2*1 = 1/2

The limit of the function, if the accumulation point is x = 0, is: lim f(x) = 1/2.

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