Use the definition of limit to prove the following limit lim x->-3 (x^2+4x+1)=-2 in epsilon delta? .

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Answer 1 · 2018-06-23T15:25:45

Please see below.

Given #epsi > 0#, let #delta = min{1,epsi/3} #

Now for any #x# with #0 < abs(x-(-3)) < delta#,

first note that #abs(x+3) < 1#,

so #-1 < x+3 < 1#,

and #-3 < x+1 < -1#.

Consequently, #abs (x+1) < 3#.

Further more, we now see that

#abs((x^2+4x+1) - (-2)) = abs(x^2+4x+3)#

# = abs((x+1)(x+3))#

# = abs(x+1) abs(x-(-3))#

# < 3abs(x-(-3))#

# < 3 delta#.

#< 3(epsilon/3)#

# = epsi#

So, if #0 < abs(x-(-3)) < delta#, then #abs((x^2+4x+1) - (-2)) < epsilon#

and, by definition,

#lim_(xrarr-3)(x^2+4x+1) = -2#