Proving $\lim_{x\to 2} \frac{x+1}{x+2}=\frac34$ using $\epsilon$-$\delta$ definition

Question

Using the definition of the limit show that

$$\lim_{x\to 2} \frac{x+1}{x+2}=\frac34$$

I understand we let $\epsilon > 0 $ and seek $\delta$ such that $0<|x-2|<\delta $, then $\left|\frac{x+1}{x+2}-\frac34\right|<\epsilon.$ But I don't know how to show that in this case.

Answer 1

Suppose $\epsilon > 0$ be given. Let $\delta = \min\{1,16\epsilon\}$. Then $0<|x-2|<1$, or $0<x-2<1$. Add $4$ to both sides to get $4<x+2<5$. Thus, $|x+2|=x+2>4$, and so $\frac 1{|x+2|}<\frac 14$. We now have \begin{align} \left|\frac{x+1}{x+2}-\frac34\right| &=\frac{|x-2|}{4|x+2|} \\ &< \frac{\delta}{4|x+2|} \\ &< \frac {\delta}{16} \\ &= \epsilon. \end{align}

(Note that the choosing of a $\delta$ is dictated at the end, after the raw scratchwork is organized. This is sort of what I did in writing this answer.)

Answer 2

hint: $\left|\dfrac{x+1}{x+2} - \dfrac{3}{4}\right| = \dfrac{|x-2|}{4|x+2|}< \dfrac{|x-2|}{4}< \epsilon$, next try to choose an initial $\delta = ??$ to make first $|x+2| > 1$ for $|x-2| < \delta$ ? note that $\delta \leq 4\epsilon$

Answer 3

Hint: Let $\epsilon \gt 0$. We want to find a $\delta \gt 0$ such that $0\lt|x-2|<\delta\implies|\dfrac{x+1}{x+2}-\dfrac{3}{4}|\lt \epsilon$. So consider

$|\dfrac{x+1}{x+2}-\dfrac{3}{4}|=\dfrac{1}{4}|\dfrac{4x-8+9}{x+2}|\le |\dfrac{x-2}{x+2}|+\dfrac{9}{4}|\dfrac{1}{x+2}|=\dfrac{1}{x+2}[|x-2|+\dfrac{9}{4}]$. Proceed from here.

Answer 4

Maybe this will make it clearer (I'll be a little bit more wordy than usual, but maybe it will help): Given $\epsilon >0$, we must find $\delta>0$ such that $$ \left|\frac{x+1}{x+2}-\frac{3}{4}\right|=\left|\frac{4x+4-3x-6}{4(x+2)}\right|=\left|\frac{x-2}{4(x+2)}\right|=\frac{1}{4}\cdot\frac{|x-2|}{|x+2|}<\epsilon $$ whenever $0<|x-2|<\delta$. Restrict $x$ to lie in the interval $|x-2|<1$, noting that $$ -1<x-2<1 \Longleftrightarrow 3<x+2<5\Longleftrightarrow \frac{1}{3}>\frac{1}{x+2}>\frac{1}{5}\Longleftrightarrow \frac{1}{12}>\frac{1}{4|x+2|}>\frac{1}{20}. $$ So choose $\delta=\min\{1,12\epsilon\}$. Now check that this choice of $\delta$ works.

Check that choice of $\delta$ works: Given $\epsilon>0$, we let $\delta=\min\{1,12\epsilon\}$. If $0<|x-2|<\delta$, then $\frac{1}{12}>\frac{1}{4|x+2|}$, as we showed above. Also, $|x-2|<12\epsilon$, so $$ \left|\frac{x+1}{x+2}-\frac{3}{4}\right|=\frac{1}{4|x+2|}\cdot |x-2|<\frac{1}{12}\cdot12\epsilon=\epsilon. $$ This shows that $$ \lim_{x\to 2}\frac{x+1}{x+2}=\frac{3}{4}. \quad\blacksquare $$

Answer 5

$$\frac{x+1}{x+2} - \frac{3}{4} = \frac{4x + 4 - 3x - 6}{4x + 8} = \frac{x - 2}{4x + 8}$$

$$\frac{|x-2|}{|4x+8|} < \frac{\delta}{|4x+8|} < \epsilon$$

Let $\delta < 1$. Then $|4x+8|$ is between $12$ and $20$. So

$$\frac{\delta}{|4x + 8|} \leq \frac{\delta}{12} < \epsilon$$

when we choose $\delta < \min(12 \epsilon, 1)$.

Answer 6

$$\left| \frac{x+1}{x+2}-\frac34 \right| = \left| \frac{x-2}{4(x+2)}\right| $$

Now, choose $0<\delta \le 1$. This implies that $|x-2|<1$, which in turn implies

$$\left| \frac{1}{4(x+2)}\right|\le \frac{1}{12}$$

Then, for this choice of $\delta$ we have

$$\left| \frac{x+1}{x+2}-\frac34 \right| < \frac{1}{12} \left| x-2 \right|$$

Now, given any $\epsilon>0$,

$$\left| \frac{x+1}{x+2}-\frac34 \right| < \frac{1}{12} \left| x-2 \right|<\epsilon $$

whenever $|x-2|<\min(1,12 \epsilon)$. Therefore, choose $\delta =\min (1,12\epsilon)$ and we have it!