Why can't i solve limits by definition like this?

Question

I want to show that as x approaches $2,\ \frac{-1}{x+2}$ approaches $\frac{-1}{4}$.

The way I solved it was:

$\forall \epsilon>0,\ \exists \delta>0$ s.t. $|x| < \delta \Rightarrow |\frac{-1}{x+2} + 1/4 |< \epsilon$

Let epsilon be fixed and delta s.t. $\frac{-1}{\delta+2} + \frac{1}{4} = \epsilon$

Some calculations give us that $\delta = \frac{-1}{(\epsilon - \frac{1}{4})} -2$ if I'm not mistaken. I've used some values of epsilon and this works, and honestly, it just makes sense.

However, my teacher solved it like this: ...$|x-2|<\delta \Rightarrow |(\frac{-1}{x+2} + \frac{1}{4})| < \epsilon$.

let $x$ be equal to $2+h$. $x$ tends to $2$ is the same as h tending to $0$. Some calculations with the function gives us $|h/(|4(4+h))| \le h/12 \le \delta/12 \le \epsilon$.

Let's impose that $\delta < 1$, and so $-1 < h < 1$ and $12 < 4(4h) < 20$. All we have to do is consider $delta = min \{1,12\epsilon\}$

My question is, is it incorrect to do it my way? It just seems so much simpler.

Answer 1

Your method is wrong because by definition of $\lim_{x \rightarrow 2} \frac{-1}{x+2}=-\frac14$,

you have to show that

$$\forall \epsilon >0, \exists \delta >0, 0<|x-2|< \delta \implies \left| \frac{-1}{x+2}+\frac14\right| < \epsilon$$

What you have tried to show is $\lim_{x \rightarrow 0} \frac{-1}{x+2}=-\frac14$

of which we know it is not true since $\lim_{x \rightarrow 0} \frac{-1}{x+2}=-\frac12$

Note that you have to show it for all $\epsilon>0$ and not just some $\epsilon>0$

Answer 2

Your approach started out in a valid way, but care must be takem. Unfortunately, $$ \frac{-1}{\epsilon - \frac14} -2 $$ is not positive for all $\epsilon$ so it will not satisfy the limit definition.