Prove function in little-oh of a big-theta function
$f(n)$ is $o( g(n) )$ 

This above equation implies the following:

$f(n) &lt; c \cdot g(n) \  -(1)$

for some $c \in R$ and $\forall n \geq n_{0}$

$g(n)$ is $\Theta( h(n) )$

The above equation implies following:

$ c_{1} \cdot h(n) \leq g(n) \leq c_{2} \cdot h(n) -(2)$

for some $c1, c2 \in R$ and $\forall n \geq n_{0}'$

Now let's assume $c$ > 0

Multiplying equation $(2)$ by $c$ gives:

$ c \cdot c_{1} \cdot h(n) \leq c \cdot g(n) \leq c \cdot c_{2} \cdot h(n)$

Considering right side of the above inequality:

$c \cdot g(n) \leq c \cdot c_{2} \cdot h(n)  -(3)$

Let $n_{0}''$ be $max(n_{0}, n_{0}')$

We can say $\forall n \geq n_{0}''$, equation $(1)$, $(2)$ and $(3)$ will hold true. Now by using $(1)$ and $(3)$, we can infer:

$ c \cdot c_{2} \cdot h(n) &gt; f(n) -(4)$

Let $c' = c \cdot c_{2}$

Therefore we can write equation $(4)$ as:

$ f(n) &lt; c' \cdot h(n)$

$\forall n \geq n_{0}''$ and since $R$ is closed under multiplication, therefore $c \in R$

which implies $f(n)$ is $o(h(n))$.

Note: For $c &lt; 0$, you can consider left side of the inequality, which after multiplying by $c$ will actually come to right side (since $c &lt; 0$).