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,

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