Proving the limit of a sequence.

[gleaming lagoon]

For a large $n$ you can say the the nominator, $n-5$, will behave like $n$ and the denominator, $25n^2+15n-50$, like $25n^2$. Then you will get
$$\left| \frac{n}{25n^2}\right| = \left| \frac{1}{25n}\right| = \frac{1}{25n} < \epsilon$$

@warm tartan

I don't know if you still needed an answer

I don't think it works on this server

[crude raft]

zlik

[gleaming lagoon]

if it's a sequence, you don't need the epsilon delta definition

for sequences this is the definition of a limit
$$\exists L \in \mathbb{R} : \forall \epsilon > 0 : \exists n_0 \in \mathbb{N} : \forall n \geq n_0 : \left| a_n - L \right| < \epsilon $$

[crude raft]

zlik

[gleaming lagoon]

so no $\delta$

[crude raft]

zlik

[gleaming lagoon]

No, I know that this is not the full proof but this is what you need to prove it

no worries

[gleaming lagoon]

How did you actually prove it?