Let f(x) and g(x) be two continuous functions and h(x)=limn→∞x2n⋅f(x)+x2m⋅g(x)(x2n+1),m∈R. If limx→1h(x) exists, then one root of f(x)−g(x)=0 is

Let f(x) and g(x) be two continuous functions and h(x)=limn→∞x2n�

1.	Let f(x) and g(x) be two continuous functions and h(x)=limn→∞x2n⋅f(x)+x2m⋅g(x)(x2n+1),m∈R. If limx→1h(x) exists, then one root of f(x)−g(x)=0 is
Answer» Let $f (x)$ and $g (x)$ be two continuous functions and $h (x) = lim n \to \infty \frac{x^{2 n} \cdot f (x) + x^{2 m} \cdot g (x)}{(x^{2 n} + 1)}, m \in R$ . If $lim x \to 1 h (x)$ exists, then one root of $f (x) - g (x) = 0$ is