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$\begingroup$

USING THE DEFINITION:

For any $ε > 0$ , there is $δ > 0$ ; $| x - a | < δ$ Then $| F(x) - L | < ε$

SHOW THAT:

$$\lim_{x \rightarrow -2}(\frac{-x}{2}-4)=-3$$

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    Please share what you have done. – 2017-02-19
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    And what else? What have you tried? – 2017-02-19
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    Start by letting Epsilon$>0$ and assuming $|x+2|<$delta. – 2017-02-19
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    | F(x) - L | < ε | ( -x/2 -4 ) + 3 | < ε | ( -x/2 + 1 ) | < ε | ( -X + 2 / 2 | < ε | ( -X + 2 ) | < 2ε Now I just want to remove the " -" before " X" but how? – 2017-02-19
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    One sec I got you with a proof – 2017-02-19
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    I need to get to |x+2| but now I am stuck here | -X + 2 | – 2017-02-19
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    Isn't $-x/2-4+3=-x/2-1$ and not $-x/2+1$? – 2017-02-19
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    thanks for help, I found the solution. – 2017-02-19
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    With absolute value you can factor out negatives, if this helped please upvote and accept as answer.:] – 2017-02-19

1 Answers 1

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$$f(x)=-\frac{x}2-4$$

The goal is to find $\delta$ such that $\delta > \lvert x-(-2)\rvert > 0 \implies \exists\epsilon>0,\,\epsilon > \lvert f(x)-(-3) \rvert$

Start with $\epsilon$ and use that to define $\delta$:

$$\epsilon > \lvert f(x)-(-3) \rvert>0$$ $$\epsilon > \lvert (-\frac{x}2-4)+3 \rvert>0$$ $$\epsilon > \lvert -\frac{x}2-1 \rvert>0$$ $$\epsilon > \lvert-\frac12\rvert\cdot\lvert x+2\rvert>0$$ $$\epsilon > \frac12\lvert x+2\rvert>0$$ $$2\epsilon > \lvert x-(-2)\rvert>0$$

Now, simply define $\delta=2\epsilon$ and the proof is complete. In order to verify this, simply work through the above steps in reverse.