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How do you find the limit of #(x+3)^2006# as #x->-4#?

Answer 1

By directly substituting in the value.

We can substitute #-4# for #x# directly. Unlike other limit problems, the function is defined at #x=-4# so we can plug it right into the function.

#lim_(xrarr-4)(x+3)^2006=(-4+3)^2006=(-1)^2006=1#

Note that #(-1)^m=1# when #m# is even and #(-1)^n=-1# when #n# is odd. You can prove this to yourself by multiplying out a couple examples like #(-1)^3#, #(-1)^4#, and #(-1)^5#. In an even-powered example, each #-1# will be paired with another #-1#, leaving a positive result. With an odd power, there will always be one leftover #-1#.

You can also think about #(-1)^2006# as #((-1)^2)^1003=1^1003=1#.

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Answer 2

William Abdalla

To find the limit of (x+3)^2006 as x approaches -4, we can substitute -4 into the expression and evaluate it.

((-4)+3)^2006 = (-1)^2006 = 1

Therefore, the limit of (x+3)^2006 as x approaches -4 is 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the limit of #(x+3)^2006# as #x-&gt;-4#?

How do you find the limit of #(x+3)^2006# as #x->-4#?