Find the limit as x approaches infinity of y=arccos((1+x^2)/(1+2x^2))?

1 Answer
Aug 18, 2014

There is a law of limits that deals with composite functions. Essentially, if we have two functions f and g:

lim_(x->a) f(g(x)) = f(lim_(x->a) g(x))

In this case, f(g(x)) would be arccos(g(x)), and g(x) would be (1+x^2)/(1+2x^2).

So, to find the limit of the entire thing as x approaches infinity, we can find the limit of the inner function as x approaches infinity, and then evaluate the outer function with this value:

lim_(x->infty) arccos((1+x^2)/(1+2x^2)) = arccos(lim_(x->infty) (1+x^2)/(1+2x^2))

It should be easy to see that lim_(x->infty) (1+x^2)/(1+2x^2) is equal to 1/2.

So, we have:

lim_(x->infty) arccos((1+x^2)/(1+2x^2)) = arccos(1/2)

The arccosine of 1/2 is equal to pi/3:

lim_(x->infty) arccos((1+x^2)/(1+2x^2)) = pi/3