Normally equations do not work like $\iff$ statements. That is, if I want to prove that $\sin 2x = 2 \sin x \cos x$ I can just start from one side and show that it's equal to the other.

However, earlier I was trying to prove that if $f$ is continuous, then 

$$\lim\_{z \to z\_0} h(f(z)) = \lim\_{z \to f(z\_0)}h(z)$$

In doing this problem, I felt that it's neccesary to first assume the limit on the RHS exists, and prove that the LHS exists and has the same value. Then assume that the one on the LHS exists, and prove the RHS has the same value.

What exactly is the difference between these two scenarios?

**Edit**
Suppose the limit on the $RHS$ exists and equals $L$.

Fix $\epsilon > 0$. Then there is a $\delta\_1$ such that $\forall z :|z - f(z\_0)| < \delta\_1 \implies |h(z)-L| < \epsilon$ 

Because $f$ is continuous, there exists a $\delta\_2$ such that $\forall z: |z-z\_0|<\delta\_2 \implies |f(z) - f(z\_0)| < \delta\_1$

Stringing these together, we have $\forall z: |z-z\_0|<\delta\_2 \implies |f(z) - f(z\_0)| <\delta\_1 \implies |h(f(z))-L| < \epsilon$.

Thus $\forall z: |z-z\_0|<\delta\_2 \implies |h(f(z))-L| < \epsilon$, so the limit on the $LHS$ exists and equals $L$.