If the limit of $y$ is $L$ would the limit of $1/ y$ yield $1/L$? For instance the  $$\lim_{x=0}\frac{\sin x}{x}=1$$ Also $$\lim_{x=0}\frac{x}{\sin x}=1$$ yields the reciprocal of 1 which is 1 would this be true for all situations?
Another example would be 
$$\lim_{x=0}\frac{1-\cos x}{x}=0$$
$$\lim_{x=0}\frac{x}{1-\cos x}=\frac{1}{0}=\text{DNE}$$
 A: If
$$\lim_{x\to a} f(x)=L,$$
then:
$$\lim_{x\to a} \frac{1}{f(x)}=\frac{1}{\lim_\limits{x\to a} f(x)}=\begin{cases} \frac{1}{L}, \ \ L\ne 0, \\ 
\begin{cases} +\infty, L=\lim_\limits{x\to 0} x^2=0, \\ -\infty, L=\lim_\limits{x\to 0} -x^2=0, \\ undefined, L=\lim_\limits{x\to 0} x\sin{\frac{1}{x}}=0
\end{cases}, \\ undefined, \ \ f(x)=0\end{cases}$$
A: In general, if $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} g(x) = M \ne 0$, then $$\lim_{x\to a} \frac{f(x)}{g(x)} = \frac{L}{M}.$$
If both $L$ and $M$ are zero, then the limit could exist, but more work is required.  In the case of $\lim_{x\to 0} \frac{\sin(x)}{x}$, we can (for example) make some clever geometric arguments to show that the limit is 1.  More broadly, L'Hospital's rule could allow us to compute the limit, provided that it actually applies.
Finally, if $L\ne 0$ and $M=0$, then the limit does not exist.  However, it could fail to exist in two distinct ways:  it could be infinite (either $\pm\infty$—for example, $\lim_{x\to 0} x^{-2} = +\infty$), or it could simply fail to exist altogether (for example,
$$ \lim_{x\to 0^+} \frac{1}{x} = +\infty
\qquad\text{and}\qquad
\lim_{x\to 0^-} \frac{1}{x} = -\infty;$$
thus
$ \lim_{x\to 0} \frac{1}{x} $
does not exit).
