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The $\epsilon$-$\delta$ concept still hasn't clicked so I am doing some really easy examples to try to get it

Prove: $\lim \limits_{x \to 2} 4x=8$.

Now, I have to prove that for all $\epsilon>0$, there exists $\delta$, such that $0 < (x-2)<\delta$ leads to $|4x-8|<\epsilon$.

Now, I know $|4x-8|=4|x-2|$, and let $x$ be not greater than $1$ distance away from $2$ (arbitrary choice), therefore, $1< x<3 \implies |x-2|<1$.

For $4|x-2| <\epsilon$ , this is the part where I get stuck.

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1 Answer 1

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Let $\delta = \epsilon/4$. Then if $\vert x - 2 \vert < \delta$, we have $$4\vert x - 2 \vert < 4 \cdot \epsilon/4 = \epsilon.$$

In general, if you have a linear function $y = mx + b$ with $m \neq 0$, you can show that $\lim_{x \to a}f(x) = ma + b$ as follows:

Pick $\epsilon > 0$. Let $\delta = \epsilon/m$. Then if $\vert x - a \vert < \delta$, we have $$\vert mx + b - (ma + b)\vert = m\vert x - a \vert < m \cdot \epsilon/m = \epsilon.$$ Note that if $m = 0$, we have $m \vert x - a \vert = 0 < \epsilon$ regardless of our choice of $\delta$.

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