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I have been trying to find the limit of following question but can't seem to get the right answer. I first took the logs because the limit of the power is undefined and then tried to solve the limit using substitution. But I keep getting $1$, when the answer converges towards $0$.

$$\lim_{x\to\infty}\left(\frac{x^3}{x^3+1}\right)^{(3x^4+2)/x}$$

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  • $\begingroup$ Please incorporate the question into the post instead of posting a link. Links tend to rot and many people don't like clicking through. Then show your work because it is much easier to give a good answer to what went wrong when we see it. $\endgroup$ – Ross Millikan Oct 16 '18 at 2:05
  • $\begingroup$ Thank you. I will keep that in mind. I took a picture because I was having trouble writing my working out procedure as I am new to Latex. $\endgroup$ – David MB Oct 16 '18 at 2:53
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When you are taking limits as $x \to \infty$ you want terms like $\frac 1x$. This should prompt you to see $\frac {x^3}{x^3+1}=1-\frac 1{x^3+1}$

Now I will work informally-you need to justify this. The $+1$ doesn't matter in $\frac 1{x^3+1}$ so we are asking about $\left(1-\frac 1{x^3}\right)^{(3x^3+\frac 2x)}$ and the $\frac 2x$ doesn't matter so we have $\left(\left(1-\frac 1{x^3}\right)^{x^3}\right)^3$ Does the inside of the outer parentheses look familiar?

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  • $\begingroup$ Thank you;I solved it. But I want to understand why you took terms out of consideration, like 2/x. Is it because it negligible? And if that the case, should I follow the same procedure of removing negligible terms in other limit calculations? $\endgroup$ – David MB Oct 16 '18 at 3:35
  • $\begingroup$ The informal answer is that terms are negligible until they matter. You need to find the leading order term that survives and it will dominate all the others. Some of these problems are carefully balanced to cancel out many high order terms, When I say a term doesn't matter it is because some other term is much larger or smaller. To be formal you should prove that term is negligible. $\endgroup$ – Ross Millikan Oct 16 '18 at 3:44
  • $\begingroup$ I see. Thank you, I will definitely remember this. $\endgroup$ – David MB Oct 16 '18 at 4:03
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Hint: Disregard the +2 in $3n^4+2$ and simplify a bit. You end up with a lot of $n^3$ terms that are going to infinity. Try a substitution like $m=n^3$.

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