How do you find the limit of #sinx^2/(1-cosx)# as #x-0#?

Question

How do you find the limit of #sinx^2/(1-cosx)# as #x->0#?

Christopher Allers · Accepted Answer

#sinx^2/(1-cosx) = sinx^2/(1-cosx) (1+cosx)/(1+cosx)# # = (sinx^2(1+cosx))/(1-cos^2x)# # = sinx^2 * 1/sin^2x  (1+cosx)# # = sinx^2/x^2 * x^2/sin^2x * (1+cosx)# Using #lim_(thetararr0)sintheta/theta#, and #lim_(xrarr0)cosx - 1# we get #lim_(xrarr0) sinx^2/x^2 * x^2/sin^2x * (1+cosx) = (1)(1)(1+1) = 2#

Charles Arocha · Answer

#=2#

The Taylor Expansions for sine and cosine are:

#sin (u) = u - u^{3}/{3!}+ \O (u^5)#

and

#cos(u) = 1 - u^2/(2!) + \O(u^4)#

Plugging these into the limit yields:

#lim_(x to 0) (x^2 - x^{6}/{3!} + \O (x^10))/(1-(1 - x^2/(2!) + \O(x^4)))#

#= lim_(x to 0) (x^2 - x^{6}/{3!} + \O (x^10))/( x^2/(2!) + \O(x^4))#

Divide top and bottom by #x^2#:

#= lim_(x to 0) (1 + \O (x^4))/( 1/(2!) + \O(x^2)) = 2#

Adam Adkins · Answer

undefined To find the limit of sinx^2/(1-cosx) as x approaches 0, we can use the concept of L'Hôpital's Rule. By applying this rule, we differentiate the numerator and denominator separately with respect to x.

Differentiating sinx^2 with respect to x gives us 2x*cosx, and differentiating 1-cosx with respect to x gives us sinx.

Now, we substitute x=0 into the derivatives we obtained. We get 2(0)*cos(0) in the numerator, which equals 0, and sin(0) in the denominator, which also equals 0.

Since both the numerator and denominator evaluate to 0, we can apply L'Hôpital's Rule again. By differentiating the numerator and denominator once more, we obtain 2*cosx - 2x*sinx in the numerator, and cosx in the denominator.

Substituting x=0 into these derivatives, we get 2*cos(0) - 2(0)*sin(0) in the numerator, which simplifies to 2, and cos(0) in the denominator, which equals 1.

Therefore, the limit of sinx^2/(1-cosx) as x approaches 0 is 2.

How do you find the limit of #sinx^2/(1-cosx)# as #x-&gt;0#?

How do you find the limit of #sinx^2/(1-cosx)# as #x->0#?