2
$\begingroup$

I'm trying to solve the problem of showing that $$\forall x \neq -1$$ $$\lim_{x\to a}\left(\frac{x}{1+x}\right) = \frac{a}{1+a}$$ using the $\epsilon$-$\delta$ definition of a limit.

I've broken it down till $$\frac{|x-a|}{|(1+x)(1+a)|}$$

I now need a lower bound for $$|1+x|$$

I realize this is probably really simple but I'm just scared of making a mistake.

Given that a can be either positive or negative I'm not sure how to pick the lower bound exactly?

$\endgroup$
  • $\begingroup$ Can't you just use the fact that both limits exists and the lower limit is never zero, therefore the limit of the divison of the functions equals the division of the limits? $\endgroup$ – Rono Nov 4 '14 at 3:33
  • 1
    $\begingroup$ Specifically says to use epsilon delta, I cant use the Limit Laws. $\endgroup$ – Exc Nov 4 '14 at 3:39
1
$\begingroup$

Note $a\neq -1$. So $|a+1|>0$. If we select $\delta<\frac{|a-(-1)|}{2}$ ( of course with other conditions related with $\epsilon$), then for $|x-a|<\delta$, we should have $$|x+1|\geq |a+1|-|a-x|>|a+1|-\frac{|a-(-1)|}{2}=\frac{|a+1|}{2}>0$$

This gives you a lower bound. The idea is that we force $x$ be closer to $a$, so the distance from $x$ and $-1$ must be larger than half of the distance between $a$ and $-1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.