<s>[INST] How to calculate $\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$ $$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$$
someone please help i’m not sure how to compute this. i’ve tried to do it this way:$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})} = \frac {1-1/7^{4x} }{1+ 1/21^{4x}} $$
$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}
= \dfrac {\lim\limits_{ x \to \infty } 1-1/7^{4x} }{ \lim\limits_{ x \to \infty } 1+ 1/21^{4x} }$$
 but that’s defiantly wrong
 [/INST] In that limes, $x\to\infty$ is equivalent to $7^{2x}\to\infty$, hence:
$$\lim\limits_{x \to \infty} \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}
= \lim\limits_{z \to \infty} \frac {z +z^{-1}}{ 3(z - z^{-1})}
= \frac13\lim\limits_{z \to \infty} \frac {z^2+1}{z^2 - 1}
= \frac13
$$
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