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

2 Answers
Mar 3, 2016

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

Explanation:

#(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#

Mar 3, 2016

Just do the calculations, without worring with limits.

Explanation:

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