Prove $\lim\limits_{n \to \infty} \frac{n^2+1}{5n^2+n+1}=\frac{1}{5}$ directly from the definition of limit.

So far ive done this:


It must be shown that for any $\epsilon>0$, there exists a positive integer $N$ such that for $n\in \mathbb{N}$ with $n\ge N$, one has

$$\left|\frac{n^2+1}{5n^2+n+1}-\frac{1}{5}\right| <\epsilon \text{ (or equivalently,}\frac{1}{5}-\epsilon \lt \frac{n^2+1}{5n^2+n+1} \lt \frac{1}{5} + \epsilon \text{).}$$

$$\begin{align} \left|\frac{n^2+1}{5n^2+n+1}-\frac{1}{5}\right| & = \left|\frac{(5(n^2+1))-(1(5n^2+n+1))}{5(5n^2+n+1)}\right| \\ & = \left|\frac{5n^2+5-5n^2-n-1}{25n^2+5n+5}\right| \\ & = \left|\frac{4-n}{25n^2+5n+5}\right| \\ \end{align}$$

How do i go about choosing the specific terms or factors of $n$ to compare against the numerator and denominator in this particular problem and i guess these type of problems in general?

Like in the answer: https://math.stackexchange.com/a/553466/227134

How do you know to setup the inequality $4n+7\le5n$ for the numerator and so forth?

I cant quite figure out what to put on the right side of $4-n\le $ ?

  • $\begingroup$ Hint: $|4-n|=|n-4|$ $\endgroup$
    – WWK
    Commented Apr 17, 2015 at 2:54
  • $\begingroup$ You can replace the denominator by $25 n^2$. $\endgroup$
    – copper.hat
    Commented Apr 17, 2015 at 2:58
  • 1
    $\begingroup$ you've missed out $n^2$ in the denominator after expanding the bracket $\endgroup$
    – JMP
    Commented Apr 17, 2015 at 3:18

1 Answer 1


$$|\frac{4-n}{25n^2+5n+5}|< ϵ\\\frac{|4-n|}{|25n^2+5n+5|}< ϵ\\\frac{|n-4|}{|25n^2+5n+5|}< ϵ\\$$ now $n∈N ,n \rightarrow \infty $so $n-4>0 ,|n-4|=n-4 $ also $25n^2+5n+5>0 \rightarrow |25n^2+5n+5|=25n^2+5n+5$ $$\frac{n-4}{25n^2+5n+5} <\frac{n-4}{25n^2+5n}<\frac{n}{25n^2+5n}=\frac{1}{5n+1} < ϵ\\5n+1 > \frac{1}{ϵ}\\5n>\frac{1}{ϵ}-1 \\n>\frac{1}{5}(\frac{1}{ϵ}-1)\\ $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .