How do you find the limit of #(cos x - 1) / sin x^2# as x approaches 0?

Question

Christopher Allers · Accepted Answer

Use the fundamental trigonometric limits and algebra.

The fundamental trigonometric limits are: #lim_(xrarr0)(cosx-1)/x = 0# and #lim_(xrarr0)sinx/x=1# If the problem is #(cosx-1)/sin(x^2)#, then use #(cosx-1)/sin(x^2) = (cosx-1)/x^2 * x^2/sin(x^2)# So the limit we seek is equal to #lim_(xrarr0)(cosx-1)/x^2# Looking for a trick that might help, we it may eventually occur to us to try mulltiplying by #(cosx+1)/(cosx+1)# to replace the subtraction that goes to #0# with an addition that goes to #2#. #(cosx-1)/x^2 (cosx+1)/(cosx+1) = (cos^2x-1)/(x^2(cosx+1))# # = (-sin^2x)/(x^2(cosx+1)) = - sinx/x * sinx/x* 1/(cosx+1)# The limit as #xrarr0# is #-(1)*(1)*1/(1+1) = -1/2# Bonus If the problem is intended to be #(cosx-1)/(sinx)^2#, then use the same trick to get: #(cosx-1)/(sinx)^2 = (cosx-1)/(sinx)^2 (cosx+1)/(cosx+1)# # = (-(sinx)^2)/((sinx)^2(cosx+1)) = (-1)/(cosx+1)#. and, again, the limit is #-1/2# (Without using fundamental trigonometric limits).

Emilia Aguilar · Answer

Alternatively, you may also use L'Hospital's Rule to still get the same answer to the limit as #-1/2#.

Since the limit is an indeterminant type of form #0/0#, we may apply L'Hospital and differentiate the top and bottom twice to eventually obtain : #lim_(x->0)(cosx-1)/(sinx^2)=lim_(x->0)((d^2/dx^2(cosx-1))/(d^2/dx^2sinx^2))# #=lim_(x->0)(-cosx)/(-4x^2sinx+2cosx^2)# #=-1/2#

Christopher Agresta · Answer

undefined To find the limit of (cos x - 1) / sin x^2 as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (-sin x) / (2x * sin x^2). Evaluating this expression as x approaches 0, we have (-sin 0) / (2 * 0 * sin 0^2), which simplifies to 0/0. Applying L'Hôpital's Rule again, we differentiate the numerator and denominator once more. The derivative of -sin x is -cos x, and the derivative of 2x * sin x^2 is 2 * sin x^2 + 2x * 2 * cos x^2 * x. Evaluating this expression as x approaches 0, we get -cos 0 / (2 * 0 * sin 0^2 + 2 * 0 * 2 * cos 0^2 * 0), which simplifies to -1/0. Since we have obtained an indeterminate form of -1/0, we cannot directly apply L'Hôpital's Rule again. However, we can simplify the expression further by using the fact that sin x / x approaches 1 as x approaches 0. Therefore, we can rewrite the expression as (-cos x) / (2 * sin x^2 / x^2). As x approaches 0, sin x^2 / x^2 approaches 1, so the limit becomes (-cos x) / 2. Evaluating this expression as x approaches 0, we have (-cos 0) / 2, which simplifies to -1/2. Therefore, the limit of (cos x - 1) / sin x^2 as x approaches 0 is -1/2.

Isaiah Allen · Answer

undefined To find the limit of (cos x - 1) / sin x^2 as x approaches 0, you can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately and then evaluating the limit can help solve this problem. After taking derivatives, you will have a new expression, which you can then evaluate directly by substituting x equals 0.