How to get $\lim_{x\to 0}\frac{x}{\sin x} =1$ from $\lim_{x\to 0}\frac{\sin x}{x} =1$? We know the identity that 
$$\lim_{x\to 0}\frac{\sin x}{x} =1$$
However in many solved examples that I was going through , I came across the identity 
$$\lim_{x\to 0}\frac{x}{\sin x} =1$$
Although  it was never formally mentioned anywhere in the text . How does the previous identity imply this? 
Does it mean that as $x$ becomes very small the value of $\sin x$ is approximately equal to the value of $x$ so the value of both of $\frac{x}{\sin x}$ and $\frac{\sin x}{x}$ tends to $1$?

You have the limit identities such as 
$$\lim (f(x) + g(x)) = \lim f(x) + \lim g(x)$$
$$\lim f(x) g(x) = (\lim f(x) )( \lim g(x))$$
and
$$\lim \frac{f(x)}{g(x))} = \frac{\lim f(x)}{\lim g(x)}$$
etc., which hold as long as the individual limits exist (and
in the last case, we need $\lim g(x) \neq 0.$)
So you have the tools to compute:
$$\lim \frac{x}{\sin x} = \lim \frac{1}{\frac{\sin x}{x} }  =\frac{\lim 1}{\lim \frac{\sin x}{x}} = \frac{1}{1}.$$
