How do you find the limit of #(sin x)^3x²# as x approaches 0?

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Answer 1 · 2016-03-03T14:43:21

Use fundamental trigonometric limit: #lim_(xrarr0)sinx/x=1# and some algebra.

#(sinx)^3 x^2 = (sinx)^3/x^3 x^3x^2#

# = (sinx/x)^3 x^5#

Because both #lim_(xrarr0) (sinx/x)^3# and #lim_(xrarr0)x^5# exist, we can use the product property of limits.

#lim_(xrarr0) ((sinx)^3 x^2)=lim_(xrarr0) (sinx/x)^3 * lim_(xrarr0)x^5#

# = 1*0=0#

Answer 2 · 2016-03-03T15:54:54

Just do the calculations, without worring with limits.

#(sin0)^3 0^2=0^3*0^2=0#