Show that the function $ f(x)=\frac{1}{x+1}$ is continuous at $x=1$ using the $\varepsilon$-$\delta$ definition of continuity 
Show that the function $ f(x)=\frac{1}{x+1}$ is continuous at $x=1$ using the $\varepsilon$-$\delta$ definition of continuity.

My initial thoughts were that this is a straight plug in of value at $x=1$, which would give me $f(1)=\frac{1}{1+1}$ .
Missing some crucial concepts?
 A: $$ |f(x) - f(1)| = |\frac {1} {x+1}- \frac 1 2| = |\frac {x - 1}{2(x + 1)}| \le \frac 1 2|x - 1|$$
This implies that for any given positive value $\epsilon$  there exists a positive value $\delta = 2\epsilon$ such that $|f(x) - f(1)| \le \epsilon$ whenever $|x - 1| \le \delta$. 
This says that the limit of $f(x)$ as $x$ approaches $1$ is $f(1)$ which implies that $f$ is continuous at $x = 1$. 
This says no matter how small $\epsilon $ is there corresponds another positive value $\delta$ such that for all values of $x$ that are at most $\delta$ distant from $1$, the values of the function $f(x)$ will be at most $\epsilon $  distant from $f(1)$. 
No matter how close you want $f(x)$ to be to $f(1)$ there are arbitrarily many values of $x$ close to 1. 
Intuitively there is no break of the sequence of the values of the function either side of $x = 1$.
None of this matters of course if you have not grasped the $\epsilon $ - $\delta$ definition for continuity. 
If you want to read about it try "Differential Calculus" by Shanti Narayan or "Calculus" by Michael Spivak. Might want to take a look at this video too..
