Use epsilon - N definition of Limit to verify the following limits

Question

I have this problem on my Real Analysis homework and I am stuck. The requirement is to use the $\epsilon - N$ definition of limit to verify.

$$ lim \frac{3n}{\sqrt{n^2+6n+2}} = 3 $$

This is my logical definition of limit that I am working with: $$ \forall \space \epsilon >0 \space \exists \space N_{\epsilon}; \space N_{\epsilon} \space \epsilon \space \mathbb{N} \space \forall \space n \space \epsilon \space \mathbb{N} \space n > N_{\epsilon} \space |\{x\}_{n} - a| < \epsilon $$

My work so far is:

$$ \{x\}_{n} = \frac{3n}{\sqrt{n^2+6n+2}}, \space a = 3 \\ \rightarrow | \frac{3n}{\sqrt{n^2+6n+2}} - 3| < \epsilon $$

My strategy is to attempt to get a simplified fraction and transitivity to prove that there is a value between my series and epsilon such that for some index $N_{\epsilon}$ the definition is upheld. I'm currently stuck trying to simplify this fraction. Maybe there is no further simplification and I then just need to think of some value. Anyways,

$$ 1. \space Let \space a = \frac{3n}{\sqrt{n^2+6n+2}} \space and \space b = 3 \\ 2. \space Then \space we \space have \space (a-b) = \frac{a^2-b^2}{a+b} \\ a-b = \frac{(\frac{3n}{\sqrt{n^2+6n+2}})^2 - 3^2}{\frac{3n}{\sqrt{n^2+6n+2}} + 3} \\ = \space \frac{\frac{9n^2}{n^2+6n+2}-9}{\frac{3n}{\sqrt{n^2+6n+2}} + 3} \\ \frac{-54n-18}{n^2+6n+2} \times \frac{\sqrt{n^2+6n+2}}{3n+3\sqrt{n^2+6n+2}} $$

This is where I get stuck. I don't think I am able to simplify this further, so at this point should I just be focusing on attempting to find a value larger than it that is simpler? What would a reasonable value look like? This fraction is messing with me a little bit. Appreciate any advice. Should I look at the triangle inequality to try to find a fraction?

Answer 1

There is a much simpler solution! Usually, such questions do not require you to do very complicated simplifications.

Possible solution: Let $\varepsilon>0$ and $N_\varepsilon>\frac{9-3\varepsilon}{\varepsilon}$. Then, for all $n>N_\varepsilon$ we get:

$\left|\frac{3n}{\sqrt{n^2+6n+2}}-3\right|=\left|\frac{3n}{\sqrt{(n+3)^2-7}}-3\right|<\left|\frac{3n}{n+3}-3\right|=\left|\frac{3n-3n-9}{n+3}\right|=\frac{9}{n+3}<\frac{9}{N_\varepsilon+3}<\frac{9}{\frac{9-3\varepsilon}{\varepsilon}+3}=\varepsilon$

Answer 2

You are correct. Given any $\epsilon>0$, your job is to find an $N$ so that if $n>N$, we have $|x_n-3|<\epsilon$. This will typically be done by choosing $N$ large enough so that $$\frac{1}{N}<\epsilon$$ or something similar. Such choice of $N$ is dependent on the problem provided. You are under no obligation to preserve the form of the $x_n$ sequence in your verification, on the contrary, you are trying to actively force a simplification. For example, since $x_n$ is an increasing sequence, hence we can make the denominator larger and increase our distance to 3, so $$|x_n-3|<\left|\frac{3n}{\sqrt{n^2+6n+36}}-3 \right| = \left|\frac{3n}{n+6}-3 \right|= \frac{18}{n+6} $$ So we can now see what N to choose. Pick $N$ so large that $$\frac{18}{N+6}<\epsilon$$

Answer 3

Alternative approach:

This answer is somewhat off-topic, given the requirement that you are supposed to use an $\epsilon-N$ approach. However, this answer does show how much easier the problem is to solve, with a different approach.

As suggested by the comment of CyclotomicField:

For all $n \in \Bbb{Z^+},$

$$n^2 < n^2 + 6n + 2 < n^2 + 6n + 9 = (n+3)^2.$$

Therefore, for all $n \in \Bbb{Z^+}$:

$$\frac{3n}{n} = \frac{3n}{\sqrt{n^2}} < \frac{3n}{\sqrt{n^2 + 6n + 2}} < \frac{3n}{\sqrt{(n+3)^2}} = \frac{3n}{n+3}. \tag1 $$

Using (1) above, per the Squeeze Theorem, it is sufficient to show that

$$\lim_{n\to\infty} \frac{3n}{n} = 3 ~~~\text{and} ~~~\lim_{n\to\infty} \frac{3n}{n+3} = 3. \tag2 $$

In (2) above, the first assertion is immediate, and the second assertion can be attacked via

$$\frac{3n}{n+3} = \frac{3n+9}{n+3} - \frac{9}{3n + 3}. $$

Therefore,

$$\lim_{n\to\infty} \frac{3n}{n+3} = \lim_{n\to\infty}\left[\frac{3n+9}{n+3} - \frac{9}{3n + 3}\right].\tag3 $$

In (3) above, you can use the fact that if

$$\lim_{n\to\infty} f(n) = A, ~\lim_{n\to\infty} g(n) = B ~: ~A, ~B ~~\text{finite} $$

Then

$$\lim_{n\to\infty} \left[f(n) - g(n)\right] = A - B.$$

Since $~3 \times (n+3) = (3n + 9),~$ clearly

$$\lim_{n\to\infty} \frac{3n+9}{n+3} = 3.$$

Further,

As $n \to \infty$, when examining the fraction $\dfrac{9}{n+3}$, the numerator is a constant and the denominator grows unbounded. So (if desired) you can easily craft an $\epsilon - N$ demonstration that

$$\lim_{n\to\infty} \frac{9}{n+3} = 0.$$