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I'm stuck trying to determine if this sequence converges and what the limit is should it converge:

$\frac{(2n-3)^{5}-25n^{4}}{(10\log_2(3n^{6}+5) + 8n^{2})(n^{3}-20)}$


Typically we were taught to divide by the highest power to yield something in the form $\frac{1}{n}$ which tends to 0, but I'm not sure how to deal with the logarithm in the denominator.

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    the result should be $4$, expand all consider the greatest Exponent of $n$ – 2017-02-25

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Hint. One may observe that, as $n \to \infty$, $$ \frac{(2n-3)^{5}-25n^{4}}{(10\log_2(3n^{6}+5) + 8n^{2})(n^{3}-20)}=\frac{(2-\frac3n)^{5}-\frac{25}n}{\big(\frac{10\log_2(3n^{6}+5)}{n^{2}} + 8\big)\big(1-\frac{20}{n^3}\big)}\sim \frac{2^5}{8}=4. $$