What is #lim_(x->0)# #(sin(-27x))/x#?

1 Answer
Jim H
Apr 27, 2016

#-27#

Explanation:

Use #lim_(thetararr0)sintheta/theta = 1# with #theta = -27x#

#lim_(xrarr0)(sin-27)/x = lim_(xrarr0)(-27(sin overbrace((-27x))^theta)/underbrace((-27x))_theta)#

#= -27lim_(xrarr0)(sin overbrace((-27x))^theta)/underbrace((-27x))_theta#

# = -27(1) = -27#

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