So I know that this for all epsilon>0, there exists an N such that for all n>N dp$ (p_n,\frac{1}{2})<epsilon $. But in this particular example I'm having difficulty finding the $N$ for which this holds.
So I want to solve for $n$ in the equation $|(n^2+n)^{1/2}-n-\frac{1}{2}|<epsilon$, but I'm getting stuck. I think I simplified to the point that $\frac{n}{(n^2+n)^{1/2}}+n<epsilon+\frac{1}{2}$ but I'm not sure where to go from there. Any ideas?
Thanks
You certainly are complicating things. Try multiplying the expression by $\frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}.$ Then you get $$\lim_{n\to\infty}\frac{\left(\sqrt{n^2+n}-n\right)\left(\sqrt{n^2+n}+n\right)}{\sqrt{n^2+n}+n},$$ which, recalling that $(a-b)(a+b)=a^2-b^2$, is the same as $$\require\cancel \lim_{n\to\infty} \frac{\cancel{n^2}+n-\cancel{n^2}}{\sqrt{n^2+n}+n}=\lim_{n\to\infty}\frac{n}{\sqrt{n^2(1+1/n)}+n}=\\ \lim_{n\to\infty}\frac{n}{n\left(\sqrt{1+1/n}+1\right)}=\lim_{n\to\infty}\frac{1}{\sqrt{1+1/n}+1}=\frac{1}{2}.$$
To prove it using the definition of limit, avoid rewriting that absolute value inequality as a three-side inequality. Instead, rewrite it as a sistem. Solving separately the two inequalities (for $-\varepsilon$ and $\varepsilon$) you'll obtain two neighbourhood of $+\infty$, and you'll be done.