- #1
azatkgz
- 186
- 0
Please,check my solution.
Use definition of a Limit to prove it.
[tex]\lim_{n\rightarrow\infty}\frac{2n^2-\sqrt{n}}{3n^2+5logn}=\frac{2}{3}[/tex]
for [tex]\forall\epsilon>0[/tex] [tex]\exists N[/tex] such that [tex]\forall n>N[/tex]
we have
[tex]|\frac{2n^2-\sqrt{n}}{3n^2+5logn}-\frac{2}{3}|<\epsilon[/tex]
[tex]\frac{\sqrt{n}}{8n^2}<|\frac{-3\sqrt{n}-10logn}{3(3n^2+5n)}|<\epsilon[/tex]
We can choose [tex]N=\frac{1}{4\epsilon^{2/3}}[/tex]
Homework Statement
Use definition of a Limit to prove it.
[tex]\lim_{n\rightarrow\infty}\frac{2n^2-\sqrt{n}}{3n^2+5logn}=\frac{2}{3}[/tex]
The Attempt at a Solution
for [tex]\forall\epsilon>0[/tex] [tex]\exists N[/tex] such that [tex]\forall n>N[/tex]
we have
[tex]|\frac{2n^2-\sqrt{n}}{3n^2+5logn}-\frac{2}{3}|<\epsilon[/tex]
[tex]\frac{\sqrt{n}}{8n^2}<|\frac{-3\sqrt{n}-10logn}{3(3n^2+5n)}|<\epsilon[/tex]
We can choose [tex]N=\frac{1}{4\epsilon^{2/3}}[/tex]