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Here's the limit: $$\lim _{n \to \infty} \left(\frac{a^{n}-b^{n}}{a^{n}+b^{n}}\right)$$

The conditions are $b>0$ and $a>0$.

I tried this with the case that $a>b$:

$$\lim _{n \to \infty} \left(\frac{1-\frac{b^{n}}{a^{n}}}{1+\frac{b^{n}}{a^{n}}}\right)$$

It gives me the result $1$.

But, in the case of $b>a$, I don't find a solution. Thanks for your attention.

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2 Answers 2

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If $b > a$, divide both the numerator and denominator by $b^n$ to get: $$\lim_{n \to \infty} \frac{\frac{a^n}{b^n}-1}{\frac{a^n}{b^n}+1}=\frac{-1}{1}=-1$$

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    $\begingroup$ oh im so dumb how didn't i think abt it omg :'( thanks dude $\endgroup$ Dec 29, 2016 at 1:17
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    $\begingroup$ @AymanErroutabi No problem! Also, I don't think you are dumb because you did a good job with the other case! $\endgroup$ Dec 29, 2016 at 1:19
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    $\begingroup$ @Euler_Salter Why so? When $a < b$, we have that $\frac{a^n}{b^n} \to 0$ as $n \to \infty$ and thus the limit is just $\frac{0-1}{0+1}$. $\endgroup$ Dec 29, 2016 at 1:22
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    $\begingroup$ Im pretty sure this is right, apologies $\endgroup$ Dec 29, 2016 at 1:22
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    $\begingroup$ oh thats right too .when u divide the second result i mentioned by $\frac{b^{n}}{a^{n}}$ u ll get the same equation u wrote .. damn maths are so fascinating and thanks btw $\endgroup$ Dec 29, 2016 at 1:25
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Hint: In case of $b>a$ you divide by $b^n$ In case of $a=b$ it is simply zero.

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  • $\begingroup$ yeah u re absolutly right man im exhausted couldnt even think abt it lol thank you $\endgroup$ Dec 29, 2016 at 1:18

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