If $\phi (x) =\displaystyle \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}}$, then

We have,

$\displaystyle \lim_{n \rightarrow \infty} x^{2n} = \left\{\begin{matrix}0,& if |x| < 1\\ \infty, & if |x| > 1 \\ 1, & if |x| = 1\end{matrix}\right.$

Thus, we have the following cases:

CASE I When $- 1 < x < 1$

In this case, we have $\displaystyle \lim_{n \rightarrow \infty} x^{2n} = 0$

$\therefore \phi (x) =\displaystyle \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}} = g(x)$

CASE II When $|x| > 1$

In this case, we have $\displaystyle \lim_{n \rightarrow \infty} \displaystyle \frac{1}{x^{2n}} = 0$

$\therefore \phi (x) = \displaystyle \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}}$

$\Rightarrow \phi (x) =\displaystyle \lim_{n \rightarrow \infty} \frac{f(x) + \displaystyle \frac{g(x)}{x^{2n}}}{1 + \displaystyle \frac{1}{x^{2n}}} = \displaystyle \frac{f(x) + 0}{1 + 0} = f(x)$

CASE III When $|x| = 1$

In this case, we have $x^{2n} = 1\Rightarrow \displaystyle \lim_{n \rightarrow \infty } x^{2n}=1$.

$\therefore \phi (x) = \displaystyle \frac{f(x) + g(x)}{2}$