Evaluate:
#lim_(x→0) (arcsinx-x)/x^3#?
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We know that, #(1): lim_(theta to 0)theta/sintheta=1#.
#(2):sin3theta=3sintheta-4sin^3theta#.
Let, #arc sinx=theta," so that, "x=sintheta#.
Also, as #x to 0, theta to 0, and sintheta to 0," as well"#.
#:."The Reqd. Lim."=lim_(x to 0)(arc sinx-x)/x^3#,
#=lim_(theta to 0)(theta-sintheta)/sin^3theta#,
#=lim(theta-sintheta)/{1/4(3sintheta-sin3theta)}......[because, (2)]#,
#=lim{4(theta-sintheta)}/(3sintheta-sin3theta)#,
#=lim{4sintheta(theta/sintheta-1)}/{sintheta(3-sin^2theta)}#,
#=lim_(theta to 0)(theta/sintheta-1)/(3-sin^2theta)#,
#=(1-1)/(3-0^2)............[because, (1)]#.
#rArr" The Reqd. Lim"=0#.
Substituting #0# in the given function we get a #0/0# indeterminate form.
We can use the L'Hospital's Rule,
#lim_(x→0) (1/sqrt(1-x^2)-1)/(3x^2)#
#lim_(x→0) ((1-sqrt(1-x^2))/sqrt(1-x^2))/(3x^2)#
#lim_(x→0) (1-sqrt(1-x^2))/(3x^2) . lim_(x -> 0) (1/sqrt(1-x^2))#
Evaluating second limit which is #1# and applying L'Hospital's Rule to first limit,
#lim_(x→0) (-(-2x)/(2sqrt(1-x^2)))/(6x)#
#lim_(x→0) 1/(6(sqrt(1-x^2))#
Substituting #0# and we get #1/6#
