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.
Best Answer
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.