Using Continuity of a Trig. Function to Rewrite It

[johnhuntsman]

I used Wolfram Alpha to evaluate:

lim tan[(2nπ)/(1 + 8n)]
n->infinity

it says that it can use the continuity of tan(n) at n = π / 4 to rewrite the aforementioned function as:

tan[lim ((2nπ)/(1 + 8n))]
n->infinity

What is it talking about? I was taught to use certain properties of trig functions as they pertain to limits to solve limits of trig. functions, but this is a bit beyond me.

P.S. I'm not using WA to do my homework or anything, I just wanted to see how one goes about solving a trig. limit like this, as I felt that it wasn't very straighforward.

 

[LCKurtz]

Using the variable $x$ in stead of $n$, look at the limit of$$
\lim_{x\rightarrow \infty}\frac {2\pi x}{1+8x}$$What does that converge to? That will show you what they are talking about.

 

[johnhuntsman]

Using the variable $x$ in stead of $n$, look at the limit of$$
\lim_{x\rightarrow \infty}\frac {2\pi x}{1+8x}$$What does that converge to? That will show you what they are talking about.

I can see that it converges to π / 4, and I can prove it be substituting some things and simplifying it that way. But I don't understand why they can just at the very beginning of the problem rewrite it in that way from the get go.

 

[LCKurtz]

When you are dealing with continuous functions, remember that, to put it loosely, "the limit of the function is the function of the limit". That is what allows you to take the limit "across" the function as in$$
\lim_{x\rightarrow a}f(\hbox{anything}) = f(\lim_{x\rightarrow a}\hbox{anything})$$as long as the inside limit works. So if you can figure out the limit of the inside part, you are home free.

 

[johnhuntsman]

Alright then. I gotcha. Thanks, I do appreciate it.