Hello I'm struggling to prove that the function ${1\over x+1}$ is continuous at the point a=0. (The function has -1 excluded from its domain). I understand for any $\varepsilon$ we must pick a $\delta$ so that $|f(a+h)-f(a)| < \varepsilon$ where $|h| < \delta$.
However I just don't see how the fact that $|h| < \delta$ means we can choose a $\delta$ so that $|f(a+h)-f(a)|$ will be less than $\varepsilon$. That only ensures that $|h+1|$ will always be greater than $|\delta+1|$.
I'm not sure what your last statement is trying to say; if $|h|<\delta$ then $|h+1|\leq |h|+1 < \delta + 1$ - so, in fact, $|h+1|$ is never greater than $\delta+1$. You have the right idea in that, given $\delta$, you need to find some bound that holds for any $h$. The idea in particular is that $$|f(a+h)-f(a)|$$ will be subject to certain bounds if we restrict $h$ to be small. That is the idea of continuity - you can always find a small enough interval so that the function behaves nicely.
Since the function is strictly decreasing, if we restrict $h$ to the closed range $[-\delta,\delta]$, then it must obtain its maximum at $h=\delta$ or $h=-\delta$ (which, respectively, minimize and maximize the inside of the absolute value). Thus, you could write that if $|h|<\delta$ then $$|f(a+h)-f(a)| < \max(f(a)-f(a+\delta),f(a-\delta)-f(a)).$$ Then this reduces to an easier problem: Choose a $\delta$ such that $\max(f(a)-f(a+\delta),f(a-\delta)-f(a))$ is the less than (or equal to) $\varepsilon$. This would imply the desired bound on $|f(a+h)-f(a)|$.