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.
Best Answer
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 $1$.