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

Aug 18, 2014

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

${\lim}_{x \to a} f \left(g \left(x\right)\right) = f \left({\lim}_{x \to a} g \left(x\right)\right)$

In this case, $f \left(g \left(x\right)\right)$ would be $\arccos \left(g \left(x\right)\right)$, and $g \left(x\right)$ would be $\frac{1 + {x}^{2}}{1 + 2 {x}^{2}}$.

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:

${\lim}_{x \to \infty} \arccos \left(\frac{1 + {x}^{2}}{1 + 2 {x}^{2}}\right) = \arccos \left({\lim}_{x \to \infty} \frac{1 + {x}^{2}}{1 + 2 {x}^{2}}\right)$

It should be easy to see that ${\lim}_{x \to \infty} \frac{1 + {x}^{2}}{1 + 2 {x}^{2}}$ is equal to $\frac{1}{2}$.

So, we have:

${\lim}_{x \to \infty} \arccos \left(\frac{1 + {x}^{2}}{1 + 2 {x}^{2}}\right) = \arccos \left(\frac{1}{2}\right)$

The arccosine of $\frac{1}{2}$ is equal to $\frac{\pi}{3}$:

${\lim}_{x \to \infty} \arccos \left(\frac{1 + {x}^{2}}{1 + 2 {x}^{2}}\right) = \frac{\pi}{3}$