# Evaluate : $$\displaystyle{\lim_{x \to \infty}}$$ $$({7x^2+1\over 5x^2-1})^{x^5\over {1-x^3}}$$

## 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) = $$7\over 5$$ &amp;  $$\displaystyle{\lim_{x \to \infty}}$$ $$\phi$$(x) $$\rightarrow$$ – $$\infty$$

$$\implies$$ $$\displaystyle{\lim_{x \to \infty}}$$ $$(f(x))^{\phi (x)}$$ = $$({7\over 5})^{-\infty}$$ = 0

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