How do you evaluate #(sin2h*sin3h) /( h^2)# as h approaches 0?

1 Answer
Jim H
Dec 2, 2016

Use #lim_(thetararr0) sintheta/theta = 1#

Explanation:

#lim_(hrarr0)(sin2h * sin3h)/h^2 = lim_(hrarr0)(sin2h)/h * (sin3h)/h#

# = lim_(hrarr0)(sin2h)/h * lim_(hrarr0)(sin3h)/h#

# = lim_(hrarr0)(2sin2h)/(2h) * lim_(hrarr0)(3sin3h)/(3h)#

# = 2lim_(hrarr0)(sin2h)/(2h) * 3 lim_(hrarr0)(sin3h)/(3h)#

Now with #theta = 2h# in the first limit and #theta = 3h# in the second, we have the limit mentioned above, so we get

# = 2(1) * 3(1) = 6#

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