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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!​

1 Answer

2 votes

Answer:


\displaystyle ( 1 )/(3)

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


( \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 {}^{} + (3)/(n) )} }{6n}

remember that ,


  • √(ab) \implies √(a) √(b)

thus we acquire:


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

reduce fraction:


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

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


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

simplify square root:


\displaystyle ( 2 )/(6)

reduce fraction:


\displaystyle ( 1 )/(3)

hence,


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

and we're done!

User Gustavo Gabriel
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