Show that the sequence is null? Show that $\frac{14n + 26}{49n^3 +14}$ is a null sequence.
So we need to show that: $\forall \epsilon > 0, \exists X \in \mathbb{R}, \forall n > X,|\frac{14n + 26}{49n^3 +14}| < \epsilon $. 
$|\frac{14n + 26}{49n^3 +14}|= \frac{14n + 26}{49n^3 +14} < \frac{14n+26}{n}$ (since $n> 49n^3 +14 \Rightarrow \frac{1}{n} > \frac{1}{49n^3 +14} $)
So, $\frac{14n+26}{n} < \frac{14n+n}{n} $ for n $\ge $ 26
Hence, n > $\frac{15}{\epsilon}$
Taking X:= Max {$\frac{15}{\epsilon}$,26} for n > X we have $|\frac{14n + 26}{49n^3 +14}| < \epsilon $.
Is this the right working? I'm unsure about when i was reducing it to get $\frac{15}{\epsilon}$?

You have made the bound too coarse.
For all $n \geq 1$ we have
$$
\frac{14n + 26}{49n^{2}+14} < \frac{14n + 26}{n^{2}} = \frac{14}{n} + \frac{26}{n^{2}};
$$
given any $\varepsilon > 0$, we have $\frac{14}{n} < \frac{\varepsilon}{2}$ if $n > \frac{28}{\varepsilon}$ and we have $\frac{26}{n^{2}} < \frac{\varepsilon}{2}$ if $n > \sqrt{\frac{52}{\varepsilon}}$; but then $n > \max \{\frac{28}{\varepsilon}, \sqrt{\frac{52}{\varepsilon}} \}$ only if 
$$
\frac{14n + 26}{49n^{2} + 14} < \varepsilon.
$$
