USE THE EPSILON-DELTA DEFINITION OF LIMITS TO PROVE THAT
lim -3x+7=10
x-(-1)
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As Showcase_22 says, you cannot prove $\displaystyle \lim_{x \rightarrow 0} -3x+ 7= 10$- it isn't true!
What you can prove is either $\displaystyle \lim_{x\rightarrow 0} -3x+ 7= 7$ or $\displaystyle \lim_{x\rightarrow -1} -3x+ 7= 10$. Is your problem either of those?
Yes indeed. There is a typo or something in your statement.
Let's make up something similar and go with that as an example.
$\displaystyle \lim_{x\to 2}[2x-7]=-3$
$\displaystyle |(2x-7)+3|=|2x-4|=2|x-2|<{\epsilon} \;\ if \;\ |x-2|<\frac{\epsilon}{2}, \;\ so \;\ {\delta}=\frac{\epsilon}{2}$
Here's another. I am just making an example up to illustrate to you as a tutorial. Hope it helps.
$\displaystyle \lim_{x\to 4}\sqrt{x}=2$
$\displaystyle |\sqrt{x}-2|=\left|\frac{\sqrt{x}-2}{1}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}\right|=\fr ac{|x-4|}{\sqrt{x}+2}$
Now, if we pull the reins back on delta and restrict it so that $\displaystyle {\delta}\leq 4$, then:
$\displaystyle |x-4|<4$
$\displaystyle 0<x<8$
$\displaystyle 0<x<\sqrt{8}$
$\displaystyle 2<\sqrt{x}+2<\sqrt{8}+2$
$\displaystyle \frac{1}{2}>\frac{1}{\sqrt{x}+2}>\frac{1}{\sqrt{8} +2}$
$\displaystyle \frac{1}{\sqrt{x}+2}<\frac{1}{2}$
$\displaystyle \frac{|x-4|}{\sqrt{x}+2}\leq \frac{1}{2}|x-4|$
Therefore, thus, and all that:
$\displaystyle |\sqrt{x}-2|<{\epsilon} \;\ if \;\ \frac{1}{2}|x-4|<{\epsilon}, \;\ or \;\ if \;\ |x-4|<2{\epsilon}, \;\ so \;\ {\delta}=min(2{\epsilon},4)$
There.....does that help finish the one you have assuming you fix the typo or what not?.