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$$\lim_{n\to\infty}\frac{x^3 \sin(x)}{\ln(1+x^4)}$$

How can i find limits from series?

Can anyone help me , please?

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There’s no $n$ in the fraction; did you mean the limit as $x$ goes to infinity? –  Brian M. Scott Dec 25 '12 at 19:55
    
@epsilon: Your attempted edit made the question harder to read and improperly substituted $lim$ for $\lim$. –  Brian M. Scott Dec 25 '12 at 19:58
    
..and yet you don't improve your accept rate... –  DonAntonio Dec 25 '12 at 20:10
    
HOW CAN I IMPROVE MY ACCEPT RATE? –  Yigit Can Dec 25 '12 at 20:10
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See the first answer to this question: it shows exactly how to accept an answer. –  Brian M. Scott Dec 25 '12 at 20:12
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up vote 3 down vote accepted

Are you sure you want the limit as $x\to\infty$? That limit clearly doesn't exist, as $x^3/\log(1+x^4)$ goes to $\infty$ and the sine oscillates. Just in case, I'm doing the limit as $x\to0$, which does exist.

Using the Taylor series for $\sin t$ and $\log(1+t)$ (both around $0$, the second one for $t\in(-1,1)$, we have $$ \frac{x^3\sin x}{\log (1+x^4)}=\frac{x^3\,\displaystyle\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}}{\displaystyle\sum_{k=0}^\infty\frac{(-1)^k(x^4)^{k+1}}{k+1}}=\frac{\,\displaystyle\sum_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k+1)!}}{\displaystyle\sum_{k=0}^\infty\frac{(-1)^kx^{4k}}{k+1}}=\frac{1+\displaystyle\sum_{k=1}^\infty\frac{(-1)^kx^{2k}}{(2k+1)!}}{1+\displaystyle\sum_{k=1}^\infty\frac{(-1)^kx^{4k}}{k+1}}\to\frac11=1 $$ For me, it is easier to visualize using Taylor polynomials of low degree: $$ \frac{x^3\sin x}{\log (1+x^4)}=\frac{x^3(x-O(x^3))}{x^4-O(x^6)}=\frac{x^4-O(x^6))}{x^4-O(x^8)}=\frac{x^4(1-O(x^2))}{x^4(1-O(x^4))}=\frac{1-O(x^2)}{1-O(x^4)}\to\frac11=1. $$

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Actually that should be $x^4 - O(x^6)$ in the numerator, not that it matters. –  Robert Israel Dec 25 '12 at 20:59
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