I understand given $\lim_{x \to a} f(x)=L$, we must show $\forall\epsilon>0, \exists\delta >0$ such that $|x-a|<\delta \to$ $|f(x)-L|<\epsilon$.

Prove $\lim_{x \to 2} \frac{x+1}{x+2}=\frac{3}{4}$ using the epsilon delta definition of the limit.

Attempt so far:

We have $|x-2|<\delta$

Now, $|f(x)-L| =|\frac{x-2}{4(x+2)}| < |\frac{x-2}{x+2} |=\epsilon$.

$|x-2| = |x+2|\epsilon$

I'm a bit stuck choosing a value for delta. Thanks.

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    $\begingroup$ You have to use the fact that $|x-2|<\delta.$ Remember that you want to express $\varepsilon$ in terms of $\delta.$ $\endgroup$ – saulspatz Mar 2 at 13:59
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    $\begingroup$ I suggest you to write your function as $1-\frac{1}{x+2}$, in this way the computation is quite easier $\endgroup$ – Gabriele Cassese Mar 2 at 14:00
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    $\begingroup$ @Programmer $\frac{x+1}{x+2}=\frac{x+2-1}{x+2}=\frac{x+2}{x+2}-\frac{1}{x+2}=1-\frac{1}{x+2}$ $\endgroup$ – Gabriele Cassese Mar 2 at 14:08
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    $\begingroup$ Remember that you need to express epsilon in terms of delta. What you need to do now is somehow get rid of the |x+2| using the fact that |x-2|<delta. Similar to finding some expression that is greater but still useful. Try this using an actual value for delta and use the definition of absolute value to try to see what's going on. $\endgroup$ – Edcookie274 Mar 2 at 14:10
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    $\begingroup$ if you want to prove a limit you need to exclude the value at the point, in this case $x=2$, that is, the correct expression is $0<|x-2|<\delta$ instead of $|x-2|<\delta$ $\endgroup$ – Masacroso Mar 2 at 14:29

Sometimes in these sorts of limits, you have to bound $\delta$ in order to get the inequality you want. Once you have a particular value for $\delta$, any smaller value will also work. So you can assume $\delta < 1$, say, which will guarantee that $x$ doesn't get too close to the asymptote at $-2$.

So knowing that $\delta<1$, we have that $1<x<3$ and so $3<x+2<5$. You can take your inequality now

$$|x-2|<|x+2|\epsilon <5\epsilon = \delta.$$

So let $\epsilon>0$ be given and set $\delta = \min(1,5\epsilon).$

This is a common trick, so it's worth learning.


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