Find $\lim x^2+2x-1=7$ as $x \rightarrow 2$ using epsilon-delta proof. 
Find $\lim x^2+2x-1=7$ as $x \rightarrow 2$  using epsilon-delta proof.

Help, I can't find how to prove this limit.
Any ideas?
 A: Below is a detailed derivation.
Observe that $(x^2+2x-1) = x^2+2x+1-2 = (x+1)^2 -2$.
Fix any $\varepsilon > 0$. To have $\lvert x^2+2x-1 - 7\rvert \leq \varepsilon$ is equivalent to having $\lvert (x+1)^2 -9\rvert \leq \varepsilon$, which (using the identity $a^2 - b^2 = (a-b)(a+b)$) is in turn equivalent to having 
$$\lvert x+1 - 3 \rvert\cdot \lvert x+1 + 3 \rvert \leq \varepsilon$$ i.e.
$$\lvert x - 2 \rvert\cdot \lvert x+4 \rvert \leq \varepsilon.$$
Now, for any $x$ such that $\lvert x -2 \rvert \leq 1$ (this choice of $1$ is arbitrary, any positive value would work -- you would have to adapt the proof accordingly. $1$ is a nice and simple constant, though), we have $1 \leq x \leq 3$, and therefore $5 \leq x+ 4 \leq 7$. So for any such $x$ satisfying $\lvert x - 2 \rvert \leq 1$ $(\dagger)$,
$$\lvert x - 2 \rvert\cdot \lvert x+4 \rvert \leq \lvert x - 2 \rvert\cdot 7.$$
To make the RHS less than $\varepsilon$, it is then sufficient to make sure that
$$
\lvert x - 2 \rvert \leq \frac{\varepsilon}{7}
$$
i.e. taking $\delta$ to be less than $\frac{\varepsilon}{7}$ should work... as long as this also implies what we supposed above ($(\dagger)$), that is we need $\delta \leq 1$ as well.
Taking $\delta \stackrel{\rm def}{=} \min\left(\frac{\varepsilon}{7},1\right)$ will then do the job.
