Solution :
Here f(x) = \({7x^2+1\over 5x^2-1}\)
\(\phi\)(x) = \({x^5\over {1-x^3}}\) = \(x^2x^3\over 1-x^3\) = \(x^2\over {1\over x^3}-1\)
\(\therefore\) \(\displaystyle{\lim_{x \to \infty}}\) f(x) = \(7\over 5\) & \(\displaystyle{\lim_{x \to \infty}}\) \(\phi\)(x) \(\rightarrow\) – \(\infty\)
\(\implies\) \(\displaystyle{\lim_{x \to \infty}}\) \((f(x))^{\phi (x)}\) = \(({7\over 5})^{-\infty}\) = 0
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