Question

Evaluate the limit of sequence at infinity.
$\displaystyle \large{ \lim_{n \to \infty } \frac{ \sqrt{4 {n}^{2} + 3n } }{6n} }$
Please show your work too. Thanks!​

Answer 1

Answer:

$\displaystyle \frac{ 1 }{3}$

Step-by-step explanation:

we would like to compute the following limit:

$\displaystyle \lim_{n \to \infty } \frac{ \sqrt{4 {n}^{2} + 3n } }{6n}$

if we compute the limit from numerator to denominator directly,we'd end up with

$\dfrac{ \infty }{ \infty }$

which is an indeterminate form so we must do it in a different way. well to compute this limit,we can consider factoring method . notice that we can factor the numerator and that yields

$\displaystyle \lim_{n \to \infty } \frac{ \sqrt{ {n}^{2} ( 4 {}^{} + \frac{3}{n} )} }{6n}$

remember that ,

• $\sqrt{ab} \implies \sqrt{a} \sqrt{b}$

thus we acquire:

$\displaystyle \lim_{n \to \infty } \frac{ {n} \sqrt{ ( 4 {}^{} + \frac{3}{n} )} }{6n}$

reduce fraction:

$\displaystyle \lim_{n \to \infty } \frac{ {} \sqrt{ 4 {}^{} + \frac{3}{n} } }{6}$

as n approaches to ∞ ,3/n approaches to 0 which yields:

$\displaystyle \frac{ \sqrt{ 4 {}^{} } }{6}$

simplify square root:

$\displaystyle \frac{ 2 }{6}$

reduce fraction:

$\displaystyle \frac{ 1 }{3}$

hence,

$\displaystyle \lim_{n \to \infty } \frac{ \sqrt{4 {n}^{2} + 3n } }{6n} = \boxed{ \frac{1}{3} }$

and we're done!