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I wanted to try my hand (try my brain?) at a limit problem where you have to restrict $\delta$ so I made this one up, but I think something is wrong! Prove:

$$ \lim_{x\to 3} \frac{1}{1+x} = \frac{1}{4} $$

We have to prove

$$ | \frac{1}{1+x} - \frac{1}{4} | < \epsilon \text{ whenever } |x-3| < \delta$$

Given epsilon, put $\delta = \min\{1, 12\epsilon \}$

Then, if $|x-3| < \delta$, then

$$ |x-3| < 1 \text{ and } |x-3| < 12\epsilon$$

Since $|x-3| < 1$, then also $12 < 4(x+1) < 20$, but then $\frac {1}{20} < \frac{1}{4(x+1)} < \frac{1}{12}$ so that

$$|\frac{1}{4(x+1)}| < \frac{1}{12} $$

so that

$$ | \frac{1}{1+x} - \frac{1}{4} | = |\frac{4 - (1+x)}{4(x+1)}| = |\frac{x-3}{4(x+1)}| = |x-3| | \frac{1}{4(x+1)} | < \frac{12\epsilon}{12} = \epsilon.$$

$\textbf{Q.E.D}$

It feels very counterintuitive that I should end up putting $\delta = 12\epsilon$, then $\delta$ is much bigger than $\epsilon$! Usually, you end up with something like $\delta = \frac{\epsilon}{...} $ which makes me think something is wrong!

Could someone verify this proof?

Thanks in advance

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    Computations are just clean :) Nothing counterintuitive for me. – 2017-01-15
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    To show that $\lim\limits_{x\to1}\frac{x}c=\frac1c$, one would end up with $\delta=c\epsilon$ hence $\delta$ "much larger than" $\epsilon$ if $c$ is large. Nothing wrong here. – 2017-01-15

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This comes from the fact that the function $f:x\mapsto \frac{1}{1+x}$ is very "flat" at point $3$.

You can evaluate the difference between $f(x)$ and $f(x')$ with mean value theorem : $f(x)-f(x')=(x-x')f'(c)$, where $c$ lies between $x$ and $x'$.

But $f'(x)=-\frac{1}{(1+x)^2}$, which around $x=3$ is near $\frac{1}{16}$. So $f(x)$ "moves" very slowly, and a large variation of $x$ result in a far smaller variation of $f(x)$.

And BTW, your demonstration is OK :-)

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    Yeah makes sense! Thank you. ;) – 2017-01-15