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$\lim_{n \rightarrow \infty}$$\Large \frac{n^{\frac{3}{4}}}{log^9 n}$

Applying L'Hopital's rule,

$\lim_{n \rightarrow \infty}$$\Large \frac{\frac{3}{4}n^{-\frac{1}{4}}}{\frac{9log^8 n}{n}}$

$\lim_{n \rightarrow \infty}$$\Large \frac{\frac{3}{4}n}{n^{\frac{1}{4}} .9log^8 n}$

$\lim_{n \rightarrow \infty}$$\Large \frac{\frac{3}{4}n^{3/4}}{9log^8 n}$

Still function is in indeterminate form, $\large \frac{\infty}{\infty}$

We have to apply L'Hopital's rule again and again,

and eventually we will get,

$\lim_{n \rightarrow \infty}$$\Large \frac{(\frac{3}{4})^9 n^{-1/4}}{\frac{9!}{n}}$

$\lim_{n \rightarrow \infty}$$\Large \frac{(\frac{3}{4})^9 n^{3/4}}{9!}$

and this limit, regardless of the constants here, will evaluate to $\infty$ because we have *n* to a positive power in the numerator.

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