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I don't really have any idea how to find $$lim_{ntoinfty}frac{(1+7n+n^3)^4}{3-14n^{12}}$$

Can someone please give me a hint on what to do here? What I'm having trouble with especially is the powers. How do I get rid of them?
Thanks in advance!

HINT

$$frac{(1+7n+n^3)^4}{3-14n^{12}}=-frac{left(n^3left(frac1{n^3}+frac7{n^2}+1right)right)^4}{14n^{12}-3}=-frac{n^{12}}{n^{12}} frac{left(frac1{n^3}+frac7{n^2}+1right)^4}{14-frac3{n^{12}}}$$

Hint:

$$frac{(1+7n+n^3)^4}{3-14n^{12}} = frac{(color{blue}{n(n^2+7)}+1)^4}{3-14n^{12}}$$

$(n(n^2+7))^4$ has $n^{12}$ as its leading term. So, you reach something in the form of

$$lim_{x to infty}frac{ax^n+…}{bx^n+…}$$

with $n$ as the highest power. Factoring by $x^n$ results in

$$lim_{x to infty}frac{a+frac{something}{x^{n-1}}+frac{something}{x^{n-2}}+…}{b+frac{something}{x^{n-1}}+frac{something}{x^{n-2}}+…}$$

What do those other terms tend to as $x to infty$?

Hint #1: As $n rightarrow infty$, $1/n rightarrow ?$

Hint #2: How fast does $1/n^k$ approach the answer in Hint #1?

Hint:

• Divide the numerator and the denominator by $n^{12}$ might help.

We have $$lim_{ntoinfty}frac{(1+7n+n^3)^4}{3-14n^{12}}{=lim_{ntoinfty}frac{(1+7n+n^3)^4}{-14n^{12}}\=-{1over 14}lim_{ntoinfty}{left({1+7n+n^3over n^3}right)^4}\=-{1over 14}lim_{nto infty}left(1+{7over n^2}+{1over n^3}right)^4\=-{1over 14}lim_{nto infty}left(1+{7over n^2}right)^4\=-{1over 14}}$$