Question #0448b

1 Answer
Jim H
Feb 17, 2017

11

Explanation:

lim_(xrarr0)sin^2(x)^2/x^4 = lim_(xrarr0) (sinx^2/x^2 * sinx^2/x^2)

= lim_(xrarr0) (sinx^2/x^2) * lim_(xrarr0)(sinx^2/x^2)

= 1*1 = 1

Note

lim_(xrarr0)(sinx^2/x^2)=1 by using the fundamental trigonometric limit

lim_(thetararr0) sintheta/theta = 1

with theta = x^2.

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