# Problem with estimating a sequence with intuition

I've frequently used “intuition” to solve limits at infinity. For example, if someone asked me what is: $$\lim_{x \to \infty} f(x) = \frac {x^5 + x^3 + x}{x^2}$$

Or a sequence that can be represented by such a function, I would quickly argue that the only thing that matters as go through higher and higher numbers for $$x$$ would be the terms with the higher powers, and so the limit becomes:

$$\lim_{x \to \infty} f(x) = \frac {x^5 }{x^2}$$

The numerator is growing at a much faster rate than the denominator, and so the function will diverge to $$+ \infty$$. But, when I was going through a problem from Paul's online notes, my intuition didn't go well with solving it by the book. The problem is:

$$\left\{ {\frac{{\ln \left( {n + 2} \right)}}{{\ln \left( {1 + 4n} \right)}}} \right\}_{n = 1}^\infty$$

Does this converge to a value? Now, if we use L'Hopital's rule, this is a easy enough problem. It converges to $$1$$.

$$\mathop {\lim }\limits_{n \to \infty } \frac{\ln ( n + 2 )}{\ln (1 + 4n )} = \mathop \lim \limits_{n \to \infty } \frac{^{1}/_{(n + 2)}}{^{4}/_{(1 + 4n)}} = \mathop \lim \limits_{n \to \infty } \frac{1 + 4n}{4( {n + 2} )} = 1$$

But, as I said, sometimes I like to do these by intuition, and what I did was this: as $$n$$ approaches infinity, the integers “$$2$$” and “$$1$$” won't matter. So the limit becomes: $$\lim_{n \to \infty} \frac { \ln(n) }{\ln(4n)}$$

The denominator is increasing at a much faster rate than the numerator, and so the limit will converge to $$0$$.

I understand that something is wrong with my intuition, and intuition is probably not a good way to solve mathematical problems — indeed, the tutorial this problem is from itself mentions that intuition can sometimes lead you astray — but I'd love to gather some insight as to where I'm going wrong.

• Base-$a$ logarithm of $b$ is the number of digits you'll get if you write down $b$ in the base-$a$ numeral system. From this intuition you can see that base-$e$ representation of $4n$ is only a constant number of digits longer than of $n$, hence the limit must equal one. (To be precise, I don't even know if there are numeral systems with non-integer bases, but this intuition can still apply. You can read more of this stuff about logarithms at arbital.com/p/logarithm/?l=3wj – CrabMan May 6 at 0:05
• Note that in your first two shown equations $\lim_{x\to\infty}$ is missing. – ComFreek May 6 at 6:45
• The introduction is completely wrong. A limit of any real-valued function $f(x)$ is a number (provided it exists), not another function! Possibly you mean a limit $\lim f(x)$ of a function $f(x)=\frac {x^5+\dots}{x^2}$...? But then also that limit is not equal to the function $x\mapsto \frac{x^5}{x^2}$, but rather those two functions have equal limits, which (after appropriate reasoning) allows us to artificially replace one function with another inside the limit expression. – CiaPan May 6 at 8:32
• You are using waaaay more brackets than you need to. – Acccumulation May 6 at 15:30

When you have$$\lim_{n\to\infty}\frac{\ln(n)}{\ln(4n)},$$the denominator doesn't increase at a much faster rate than the numerator. As a matter of fact, since we have$$(\forall n\in\mathbb N):\ln(4n)=\ln(4)+\ln(n),$$they increase at the same rate. And now it is easy to see that the limit is indeed $$1$$.
• And when you play around with logarithms enough (for example in computer science), it will become absolutely intuitive that $\ln(n)$ and $\ln(4n)$ increase at the same rate. – JiK May 5 at 19:14
Note that $$\ln(4n)=\ln(4)+\ln n$$ so that this function grows at the same rate as $$\ln n$$. In particular $$\lim_{n\to \infty} \frac{\ln(n)}{\ln(4n)}= \lim_{n\to \infty} \frac{\ln(n)}{\ln(4)+\ln(n)}=\lim_{n\to \infty} \frac{1}{\frac{\ln(4)}{\ln (n)}+1}=1.$$ Also note that your intuition can be formalized. If $$a_n\sim b_n$$ and $$c_n\sim d_n$$ (where $$\sim$$ means that the ratio of the two sides goes to one), then $$\lim_{n\to \infty}\frac{a_n}{c_n}=\lim_{n\to\infty}\frac{b_n}{d_n}.$$