How do you find the limit of #(x+3)^2007# as #x->-4#?

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Answer 1 · 2016-11-25T18:13:48

# lim_(x rarr -4) (x+3)^2007= -1 #

Let #f(x)=(x+3)^2007#,

Then #f(x)# is a continuous function so,

#lim_(x rarr c)=f(c) AA c in RR#

# :. lim_(x rarr -4) f(x) = f(-1) = (-1)^2007 #

Now #(-1)^(2n) = +1#, and #(-1)^(2n+1) = -1 AA n in NN#

Hence #(-1)^2007 = -1#, and so:

# lim_(x rarr -4) (x+3)^2007= -1 #