Confusion with a limit of Cothx I am a Calc 2 student and am having trouble seeing this limit, the assignment is to use the definition of the hyperbolic functions to find the limit. 
$\lim_{x\to 0+} Coth(x)$
the answer is $+\infty$ 
If I work it out I get to this point: $lim_{x\ to 0+}$ $(e^(2x)+1)/(e^(2x) -1)$
Please explain it using the definition of the Hyperbolic functions. 
Work:
$lim_{x\ to 0+}$ $cosh(x)/sinh(x)$ = $lim_{x\ to 0+}$ $e^(x)+e^(-x))/e^(x)-e^(-x)$ 
$lim_{x\ to 0+}$ $e^x(e^(x)+e^(-x))/e^(x)-e^(-x))$
$lim_{x\ to 0+}$ $e^(2x)+1)/e^(2x)-1$ = (1+1)/(1-1)  = 2/0
I don't see how 2/0 relates to 1/x like everyone keeps saying, I am extremely frustrated at this point as my professor couldn't explain it without using the graph. If i knew what the graph looked like this would be pointless. Please I am missing some fundamental idea of limits or something.
 A: First remark : $\lim_{x \to 0^+}\cosh (x)=1$ and not $2$ as you wrote (typo, I guess).
You properly wrote $$\coth(x)=\frac{e^{2x}+1}{e^{2x}-1}$$ Now, remember that, for small $y$, $$e^y=1+y+O\left(y^2\right)$$ so $$e^{2x}=1+2 x+O\left(x^2\right)$$ and then, for small $x$ $$\coth(x)=\frac{1+2 x+O\left(x^2\right)+1}{1+2 x+O\left(x^3\right)-1}=\frac{2+2 x+O\left(x^2\right)}{2 x+O\left(x^2\right)}\sim \frac 1x$$
A: Observe that, for any $a>0$,
$$
\lim_{x \to 0^+}\frac{a}{x}=+\infty
$$ giving
$$
\lim_{x \to 0^+}\coth x=\lim_{x \to 0^+}\frac{1}{x}=+\infty.
$$
A: So from this point -> $\lim_{x \to 0+}  (e^{2x}+1)/(e^{2x}−1)$
If you consider the top and bottom bits separately, the top one is always going to be almost 2, whereas the bottom one will be an infinitely small number greater than $0 - e^{2x}$ for an infinitely small $x$ is just over $1$ (as $e^{2x}$ for $x=0$ is $1$). Hence you get $2/0+$, which is +∞.
Note: if $x$ goes to $0-$, you get $e^{2x}$ being just under $1$, so you would have the limit being $2/0-$, which is $-∞$.
