How do you find the Limit of $2 x + 5$ $2x+5$ as $x \to - 3$ $x->-3$ and then use the epsilon delta definition to prove that the limit is L?

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Answer 1 · 2016-11-15T19:29:16

If $x$ $x$ is very close to $- 3$ $-3$ ,

then $2 x$ $2x$ is very close to $2 (- 3) = - 6$ $2(-3) = -6$

and $2 x + 5$ $2x+5$ is very close to $- 6 + 5 = - 1$ $-6+5 = -1$ .

So $lim_(xrarr-3)(2x+5) = -1$

Proof:

Given a positive number, $epsilon$ , let $delta = epsi/2$ .

Then for every $x$ with $0 < abs (x-(-3)) < delta$ , we have $abs(x+3) < epsi/2$ .

Therefore $abs((2x+5)-(-1))$ which is equal to $abs(2x+6)$ is equal to $2abs(x+3)$ is less than $2(delta)$ which is $2(epsi/2)$ or just $epsi$ .

Writing it in mathematics we write:

For $0 < abs (x-(-3)) < delta$ , we have

$abs((2x+5)-(-1)) = abs(2x+6)$

$= 2abs(x+3)$

$< 2(delta)$

$= 2(epsi/2) = epsi$ .

So, by the definition of limit, $lim_(xrarr-3)(2x+5) = -1$

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