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$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$$ 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

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    $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Feb 12 at 23:37
  • $\begingroup$ $\lim x→∞ \frac{7^2x +7^-2x}{3(7^2x - 7^-2x)} $ sorry if this doesn’t work i’m on a phone don’t have a laptop $\endgroup$ – dolphin Feb 12 at 23:48
  • $\begingroup$ Write $\lim_{x\to\infty} \frac{7^{2x} +7^{-2x}}{3(7^{2x }- 7^{-2x})}$ for $\lim_{x\to\infty} \frac{7^{2x} +7^{-2x}}{3(7^{2x }- 7^{-2x})}$. $\endgroup$ – Shaun Feb 12 at 23:51
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    $\begingroup$ What's with all these people with limit questions on their phone? $\endgroup$ – Gae. S. Feb 12 at 23:51
  • $\begingroup$ Also, please edit your question accordingly. $\endgroup$ – Shaun Feb 12 at 23:52
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$$\begin{align} \frac13 \frac{7^{2x} + 7^{-2x}}{7^{2x} - 7^{-2x}} &= \frac13 \frac{7^{2x}(7^{2x} + 7^{-2x})}{7^{2x}(7^{2x} - 7^{-2x})} \\ &= \frac13 \frac{7^{4x} + 1}{7^{4x} - 1} \\ &= \frac13 \frac{(7^{4x} -1) + 2}{7^{4x} - 1} \\ &= \frac13 \Big(1 + \frac{2}{7^{4x} - 1}\Big) \\ &= \frac13 + \frac{2}{3(7^{4x} - 1)} \end{align}$$ Now, what happens if $x$ goes to infinity?

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  • $\begingroup$ wow this is really confusing $\endgroup$ – dolphin Feb 12 at 23:50
  • $\begingroup$ You got it? :)) $\endgroup$ – Azif00 Feb 12 at 23:51
  • $\begingroup$ why is the answer not lim𝑥→∞[1/3+2⋅7^−2𝑥/3(7^2𝑥−7^−2𝑥)] $\endgroup$ – dolphin Feb 13 at 0:02
  • $\begingroup$ I edited my answer. But yeah, it is the same. $\endgroup$ – Azif00 Feb 13 at 0:12
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Put the constant outside of the limit: $$ \begin {align} l=&\lim\limits_{ x \to \infty }\color{blue}{\frac 1 3} \frac {7^{2x} +7^{-2x} }{ (7^{2x} - 7^{-2x})} \\ l=&\color{blue}{ \frac 1 3}\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 7^{2x} - 7^{-2x}} \\ l =& \frac 1 3\lim\limits_{ x \to \infty } \frac {\color{blue}{7^{2x}}( 1+ 7^{-4x}) }{ \color{blue}{7^{2x}} (1 - 7^{-4x})} \\ l =& \frac 1 3\lim\limits_{ x \to \infty } \frac {\color{blue}{} 1+ 7^{-4x} }{ \color{blue}{} 1 - 7^{-4x}} \\ l=&\frac 1 3 \end{align} $$

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Remember that $$\coth x = \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}, \quad\implies\quad \frac{7^{2 x}+7^{-2 x}}{7^{2 x}-7^{-2 x}} = \frac{e^{(2\ln 7)x}+e^{-(2\ln 7)x}}{e^{(2\ln 7) x}-e^{-(2\ln 7) x}}=\coth((2\ln7)x). $$ Knowing the behavior of $\coth$ at infinity, that is $$ \lim_{x\to\infty} \coth x = \lim_{x\to\infty} \frac{1}{\tanh x} = 1, $$ you immediately get $L = 1/3$.

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  • $\begingroup$ +1 for the original answer. $\endgroup$ – Aryadeva Feb 13 at 4:57
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    $\begingroup$ @Isham Thank you! $\endgroup$ – giobrach Feb 13 at 8:55
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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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