- #1

lordofpi

- 17

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*Fundamentals of Calculus*, 9th), even quadratically, but for some reason I am just getting stuck on what is probably a simple problem.

## Homework Statement

Find [itex]L[/itex]. Then find ε > 0 and δ > 0 to satisfy the definition of a limit.

Given: [tex]\lim_{x \to 2} 5 - 3x [/tex]

**2. The attempt at a solution**

First I determine [itex]\lvert x-2 \rvert < \delta[/itex].

Then, I solved for the limit [itex]L[/itex] analytically

[tex]\lim_{x \to 2} 5 - 3x = 5 - 3(2) = -1[/tex]

Now given that [itex]f(x) - L < \epsilon[/itex],

this limit can ben stated as [tex] \lvert 5 -3x -(-1) \rvert < \epsilon[/tex]

Which can be restated as [tex]\lvert -3(x-2)\rvert < \epsilon[/tex]

Or even [tex]\lvert -3 \rvert \cdot \lvert x-2 \rvert < \epsilon [/tex]

So now what do I do? I am aware that [itex]\lvert -3 \rvert = 3[/itex], but to jump to that just to get an easy [itex] \frac{\epsilon}{3}[/itex] value for [itex]\delta[/itex] just seems improper. Plus I don't know how one can properly reintroduce the minus sign back into the absolute value when doing the final proof (not shown). Any thoughts on what I am overlooking?