Epsilon and Delta proof for limit $\lim_{x\to-1} \frac{x+5}{2x+3}=4$ I have a function $f(x)=\frac{x+5}{2x+3}$ and want to show that the limit
$$\lim_{x\to -1} f(x) = 4$$
is true using $\epsilon$-$\delta$. I have a trouble finding $\delta$ so that $f(x)-4$ is less than every $\epsilon$.
Can anyone help?

Given $\epsilon > 0$, for $x \in (-1.2, -0.8)$, $|2x + 3| > 0.6$. Take $\delta = \min(0.2, \epsilon/12)$, then for all $|x - (-1)| < \delta$, we have
\begin{align}
& \left|f(x) - 4\right| \\
= & \left|\frac{x + 5}{2x + 3} - 4\right| \\
= & \left|\frac{-7x - 7}{2x + 3}\right| \\
= & 7\left|\frac{x + 1}{2x + 3}\right| \\
\leq & 7|x + 1|/0.6 \\
< & 12 |x + 1| < \epsilon.
\end{align}
