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How do you evaluate the limit #(sin^2x)/(xcosx)# as x approaches #0#?

Answer 1

#lim_(x to 0) (sin^2x)/(xcosx)#

#lim_(z to 0) (sin z)/z = 1# is a well known limit so we aim to find that in the mix

#= lim_(x to 0) (sin x)/x* tan x #

we know each of these limits so the limit of the product is the product of the limits #= lim_(x to 0) (sin x)/x * lim_(x to 0) tan x#

#= 1 * 0 = 0#

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Answer 2

Charles Anctil

To evaluate the limit of (sin^2x)/(xcosx) as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get 2sinxcosx - sin^2x / cosx - xsinx. Plugging in x=0, we have 0/1, which equals 0. Therefore, the limit of (sin^2x)/(xcosx) as x approaches 0 is 0.

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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.