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Expert: - 11/17/2008


Question
find: lim x-> 0 (sin^2(5x)/x^2)

Answer
Note that this is the same as
[lim (x->0) (sin(5x)/x) ]^2.  Once this has been done, note that the top and the bottom of the fraction need to be multiplied by 5 so that we have [lim (x->0) (5sin(5x)/(5x)) ]^2.  Note that the limit can be taken on just the lim(x->0) sin(5x)/(5x).  We'll worry about that fact that this needs to be multiplied by 5 and then squared when all done.

Let u=5x, then we have the limit(u->0)sin(u)/u.  This is known to be 1.

Putting this back into the problem, we have [5*1]² = 25.

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