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How do you find the limit of #(x(1-cosx))/tan^3x# as #x->0#?

Answer 1

Isaiah Allen

#(x(1-cosx))/tan^3x# as #xrarr0#

Substitution yields the indeterminate form #0/0#

#(x(1-cosx))/tan^3x = (x(1-cosx)cos^3x)/(sin^3x)#

# = x/sinx * (1-cosx)/sin^2x * cos^3x#

The limit that is still a problem is the middle limit.

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

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

# = sin^2x/(sin^2x(1+cosx))#

# = 1/(1+cosx)#

#lim_(xrarr0)(x(1-cosx))/tan^3x = lim_(xrarr0)(x/sinx * 1/(1+cosx) * cos^3x)#

# = (1)(1/(1+1))(1^3) = 1/2#

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Answer 2

Elijah Abram

To find the limit of (x(1-cosx))/tan^3x as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (1-cosx + xsinx)/(3tan^2xsec^2x). Evaluating this expression as x approaches 0, we find that the limit is 1/3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the limit of #(x(1-cosx))/tan^3x# as #x-&gt;0#?

How do you find the limit of #(x(1-cosx))/tan^3x# as #x->0#?