How can I find the limits of the following sequence?
$$\lim _{x\to \infty }\left(\frac{7x^2-1}{7x^2+5}\right)^x$$
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1$\begingroup$ thank you for adding it for me! $\endgroup$– RastezzyMar 19, 2017 at 19:21
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2 Answers
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\begin{align} \left(\frac{7x^2-1}{7x^2+5}\right)^x &= \left(1 - \frac{6}{7x^2+5}\right)^x \\ &= \left(\left(1 - \frac{6}{7x^2+5}\right)^{\frac{7x^2+5}{6}}\right)^{\frac{6x}{7x^2+5}} \end{align} Using $\left(1-\frac{1}{y}\right)^y \to e^{-1}$ as $y \to \infty$, the inner term $\left(1 - \frac{6}{7x^2+5}\right)^{\frac{7x^2+5}{6}}$ gets close to $e^{-1}$ as $x \to \infty$. Since the outer exponent $\frac{6x}{7x^2+5}$ tends to $0$, the overall limit is $1$.
answered Mar 19, 2017 at 19:28
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$\begingroup$ Thanks for you help angryavian. One quick question, for example $limx→∞((5x-2)/(5x-4))^x^2$ would it change the way of solving having x^2? $\endgroup$– RastezzyMar 19, 2017 at 21:36
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$$\frac{7x^2-1}{7x^2+5}=1-\frac6{7x^2+5}\implies$$
$$\left(\frac{7x^2-1}{7x^2+5}\right)^x=\left[\left(1-\frac6{7x^2+5}\right)^{7x^2+5}\right]^{\frac x{7x^2+5}}\xrightarrow[x\to\infty]{}\left(e^{-6}\right)^0=1$$
answered Mar 19, 2017 at 19:27