Find $\lim_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$ How to calculate the following limit?
$$\lim\limits_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$$
$$\lim_{x\to \infty} \Big(\frac{x^2+1}{x^2-1}\Big)^{x^2}=\lim_{x\to \infty} \Big(1+\frac{2}{x^2-1}\Big)^{\frac{x^2-1}{2}\cdot\frac{2x^2}{x^2-1}}=e^{\lim_{x\to \infty}\frac{2x^2}{x^2-1}}=e^2$$
The qty in bracket tends to 1 as x→infinte and power tends to infinity u can easily prove that Lt(x→a){f(x)}^(g(x)) if f(a)→1 and g(a)→∞ then its equal to limit of e^(f(x)-1)(g(x)) as x→a so here it is.. e^(2/(x^2-1))(x^2) limit as x→ ∞ giving e^2 .. !