recategorized by
573 views

1 Answer

4 votes
4 votes

$\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.

Related questions

0 votes
0 votes
1 answer
1
MIRIYALA JEEVAN KUMA asked Oct 14, 2018
532 views
Find the limit$\operatorname { lit } _ { x \rightarrow 1 } \left\{ \left( \frac { 1 + x } { 2 + x } \right) ^ { \left( \frac { 1 - \sqrt { x } } { 1 - x } \right) } \righ...
6 votes
6 votes
1 answer
2
sumit chakraborty asked Jan 26, 2018
660 views
The value of $\lim_{x\rightarrow \infty }\left ( \frac{4^{x+2} + 3^{x}}{4^{x-2}} \right )$ is ____________
3 votes
3 votes
1 answer
3
vishal chugh asked Jan 22, 2018
559 views
3 votes
3 votes
1 answer
4
Nymeria asked Jan 21, 2018
580 views
Answer is 1. I also know the procedure. My question is why isn't this formula working?Am I doing something wrong?