Epsilon and Delta proof for limit $\lim_{x\to-1} \frac{x+5}{2x+3}=4$

Question

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?

Answer 1

Hint: We want to find a bound of the form

$$\left| \frac{x + 5}{2x + 3} - 4 \right| \leq M |x+1|$$

Start with analyzing

$$\left|\frac{x + 5}{2x + 3} - 4\right| = \left| \frac{7(x+1)}{2x + 3}\right|$$

Now put a bound on $\displaystyle \left|\frac{7}{2x + 3}\right|$ for $x$ in some interval around $-1$. There's no canonical choice for such an interval and the bound can be any finite, positive number.

Here's one option: for $|x+1| < 1/4$, that is $-5/4 < x < -3/4$, we have $$-5/2 + 3 < 2x + 3 < -3/2 + 3$$ and thus

$$|x+1| < \frac 14 \quad \Longrightarrow \quad \frac 23 < \frac{1}{2x + 3} < 2 \quad \Longrightarrow \quad \left|\frac{7}{2x + 3}\right| < 14$$

In other words,

$$|x+1| < \frac 14 \quad \Longrightarrow \quad \left| \frac{x + 5}{2x + 3} - 4 \right| < 14 \ |x+1| \quad - \color{red}{(*)}$$

Can you finish from here?

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Added: The final choice of $\delta$ is not $1/4$.

The basic proof strategy is this:

If we can establish a bound

$$\left| \frac{x + 5}{2x + 3} - 4 \right| \leq M |x+1| \quad\text{ for some } M > 0$$

then given an arbitrary $\epsilon > 0$ choosing $\delta = \epsilon/M$ enables us to write the target deduction of

$$0 < |x + 1| < \delta \quad \Longrightarrow \quad \left| \frac{x + 5}{2x + 3} - 4 \right| \leq M |x+1| < M \cdot\frac\epsilon M = \epsilon $$

The wrinkle in this plan is that such a bound does not exist for all $x$; the expression blows up in any neighborhood of $-3/2$. However what we have now shown with $\color{red}{(*)}$ that there does exist such a bound if we start with $|x+1| < 1/4$. (The idea was to constrain $x$ away from such 'blow up' points and to center it around the value of $x = -1$. This is a standard trick in $\epsilon$-$\delta$ proofs; look again at the proof that $\lim_{x\to a} 1/x = 1/a$.)

Hence we modify the strategy and can finish the proof this way: given an arbitrary $\epsilon > 0$, choose $\displaystyle \delta = \min\left( \frac 14, \frac \epsilon{14}\right)$. With that choice of $\delta$ and by $ \color{red}{(*)}$,

$$0 < |x+1| < \delta \ \Rightarrow \ |x+1| < \frac 14 \ \Rightarrow \ \left| \frac{x + 5}{2x + 3} - 4 \right| < 14 \ |x+1|$$

We also have

$$0 < |x+1| < \delta \ \Rightarrow \ |x+1| < \frac\epsilon{14}$$

These last two deductions together imply that

$$0 < |x+1| < \delta \ \Rightarrow \ \left| \frac{x + 5}{2x + 3} - 4 \right| < 14 |x+1| < 14 \cdot \frac\epsilon{14} = \epsilon$$

as desired.

Answer 2

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}

Answer 3

The algebra is easier to follow if we put $x=d-1$ and have $ d\to 0$. We have $$f(x)-4=(x+5)/(2 x+3)-4=(d+4)/(2 d+1)-4=-(7 d)/(2 d+1).$$ To keep the denominator $2 d+1$ away from $0$, observe that $$|d|<1/4\implies |2 d+1|>1/2\implies |1/(2 d+1)|<2.$$ Hence $$(|x- (-1)|=|d|<1/4)\implies |f(x)-4|=|(7 d)/(2 d+1)|\leq |(7 d)(2)|=14|d|.$$ So, given $\epsilon>0$, let $\delta=\min (1/4,\epsilon/14)$.Now we have $$|x-(-1)|<\delta \implies |x-(-1)|=|d|<1/4\implies$$ $$ |f(x)-4|\leq 14 |d|<14 \delta\leq 14(\epsilon/14)=\epsilon.$$