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

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Answer 1 · 2017-02-08T19:00:02

#=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#

Answer 2 · 2017-02-08T19:57:04

#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#