Question

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{3 n^{2}}{4 n^{2}+1}=\frac{3}{4}$$

Short answer

Question: Using the formal definition of the limit, prove that $$\lim _{n \rightarrow \infty} \frac{3 n^{2}}{4 n^{2}+1}=\frac{3}{4}$$. Answer: To prove the limit using the formal definition, we need to find a positive integer \(N\) such that for any \(\epsilon > 0\), it holds that for all \(n > N\), $$\left|\frac{3 n^{2}}{4 n^{2}+1} - \frac{3}{4}\right| < \epsilon$$. After simplifying the expression within the absolute value, we found that $$N = \left\lceil\sqrt{\frac{3}{16\epsilon}}\right\rceil$$ satisfies the given condition. Therefore, by the formal definition of the limit, we have proved that $$\lim _{n \rightarrow \infty} \frac{3 n^{2}}{4 n^{2}+1}=\frac{3}{4}$$.

Step-by-step solution

Rewrite the expression

Start by rewriting the expression as follows: $$\frac{3 n^{2}}{4 n^{2}+1} - \frac{3}{4} = \frac{3n^2 - (3/4)(4n^2+1)}{4n^2+1}$$ The goal now is to find \(N\) such that the absolute value of the above expression is less than \(\epsilon\) for all \(n > N\).

Simplify the expression

Simplify the expression in the absolute value as follows: $$\frac{3n^2 - (3/4)(4n^2+1)}{4n^2+1} = \frac{3n^2 - 3n^2 - 3/4}{4n^2+1} = \frac{-3/4}{4n^2+1}$$

Handle the absolute value

Take the absolute value of the expression: $$\left|\frac{-3/4}{4n^2+1}\right| = \frac{3/4}{4n^2+1}$$

Bound the expression

We want to find an \(N\) such that for all \(n > N\) $$\frac{3/4}{4n^2+1} < \epsilon$$ Notice that \(4n^2+1 > 4n^2\) for all \(n\). Therefore, we can write $$\frac{3/4}{4n^2+1} < \frac{3/4}{4n^2}$$ Now, we want to find an \(N\) such that for all \(n > N\), $$\frac{3/4}{4n^2} < \epsilon$$

Find the value of N

We can now solve for \(N\) as follows: $$\frac{3/4}{4n^2} < \epsilon \Rightarrow n^2 > \frac{3/4}{4\epsilon} \Rightarrow n > \sqrt{\frac{3}{16\epsilon}}$$ Let $$N = \left\lceil\sqrt{\frac{3}{16\epsilon}}\right\rceil$$ where \(\lceil x \rceil\) denotes the smallest integer greater than or equal to \(x\).

Conclude the proof

Finally, we can conclude the proof by saying that for any \(\epsilon > 0\), we have found a positive integer \(N\) such that for all \(n > N\), we have $$\left|\frac{3 n^{2}}{4 n^{2}+1} - \frac{3}{4}\right| < \epsilon$$ Hence, using the formal definition of the limit, we have proved that $$\lim _{n \rightarrow \infty} \frac{3 n^{2}}{4 n^{2}+1}=\frac{3}{4}$$