$\lim_{n \to \infty} \dfrac{1-\frac{1}{x^{2n}}}{1+\frac{1}{x^{2n}}}$
$\therefore$ When $x=1$, The numerator is $0$ , and hence $\lim=0$
When $x>1$, the numerator and denominator is $1$, and $\lim=1$
But my question is: What about when $0
$\lim_{n \to \infty} \dfrac{1-x^{-2n}}{1+x^{-2n}}$ $,x>0$
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limits
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3I wonder why the downvote. Just because it may be obvious to some, that's no good reason to downvote anonymously, rather than posting a hint or an answer. – 2017-02-11
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0Duplicate: [$\lim\limits_{n\to \infty} \dfrac{1-x^{-2n}}{1+x^{-2n}},\,x\gt0$](https://math.stackexchange.com/q/1136169/201168). (*Found using [Approach0.xyz](https://approach0.xyz/search/?q=%24\lim_{n%20\to%20\infty}%20\dfrac{1-x^{-2n}}{1%2Bx^{-2n}}%24&p=1)*) – 2017-02-12
1 Answers
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You have: $\dfrac{x^{2n}-1}{x^{2n}+1}\to -1$, because $x^{2n} \to 0$.