# How do you use the epsilon delta definition to find the limit of (x+5)/(2x+3)  as x approaches -1?

Jan 23, 2017

${\lim}_{x \to - 1} \frac{x + 5}{2 x + 3} = 4$

#### Explanation:

Write $x$ as:

$x = - 1 + \xi = \xi - 1$

and evaluate:

$\left\mid f \left(x\right) - L \right\mid = \left\mid \frac{\xi - 1 + 5}{2 \left(\xi - 1\right) + 3} - L \right\mid = \left\mid \frac{\xi + 4}{2 \xi + 1} - L \right\mid = \left\mid \frac{\xi + 4 - 2 \xi L - L}{2 \xi + 1} \right\mid = \left\mid \frac{\xi \left(1 - 2 L\right) + \left(4 - L\right)}{2 \xi + 1} \right\mid \le \left\mid \frac{\xi \left(1 - 2 L\right)}{2 \xi + 1} \right\mid + \left\mid \frac{4 - L}{2 \xi + 1} \right\mid$

We see that if we set $L = 4$ the second term vanishes and:

$\left\mid f \left(x\right) - 4 \right\mid \le \left\mid \frac{- 7 \xi}{2 \xi + 1} \right\mid = \frac{7 \left\mid \xi \right\mid}{\left\mid 1 + 2 \xi \right\mid}$

So for any given $\epsilon > 0$ we can choose ${\delta}_{\epsilon} > 0$ such that

${\delta}_{\epsilon} < \min \left(\frac{1}{2} , \frac{\epsilon}{7}\right)$

Now for $\left\mid \xi \right\mid < {\delta}_{\epsilon}$ we have:

$\xi > - \frac{1}{2} \implies \left\mid 1 + 2 x \right\mid > 1$

so that:

$\left\mid f \left(x\right) - 4 \right\mid \le \frac{7 \left\mid \xi \right\mid}{\left\mid 1 + 2 \xi \right\mid} < 7 \left\mid \xi \right\mid$

$\left\mid \xi \right\mid < \frac{\epsilon}{7} \implies \left\mid f \left(x\right) - 4 \right\mid \le 7 \left\mid \xi \right\mid < \epsilon$

and we have proved that:

${\lim}_{x \to - 1} \frac{x + 5}{2 x + 3} = 4$