Question

$\displaystyle \sf \red{ \lim_{x \to \infty} \frac{ {3x}^{5} + 21 {x}^{3} }{7 {x}^{5} - 18x + 4 } }$​​

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \lim_{x \to \infty } \: \dfrac{3 {x}^{5} + 21 {x}^{3} }{7 {x}^{5} - 18x + 4 }$

$\displaystyle = \lim_{x \to \infty } \: \dfrac{ {x}^{5} \left( 3 + \dfrac{21}{ {x}^{2} } \right) }{ {x}^{5} \left( 7 - \dfrac{ 18}{ {x}^{4} } + \dfrac{ 4}{ {x}^{5} } \right) }$

$\displaystyle = \lim_{x \to \infty } \: \dfrac{ 3 + \dfrac{21}{ {x}^{2} } }{ 7 - \dfrac{ 18}{ {x}^{4} } + \dfrac{ 4}{ {x}^{5} } }$

$\displaystyle = \dfrac{ 3 + 0 }{ 7 - 0+ 0 }$

$\displaystyle = \dfrac{ 3 }{ 7 }$

Answer 2

Answer:

We have,

$\displaystyle \lim_{x \to \infty } \: \dfrac{3 {x}^{5} + 21 {x}^{3} }{7 {x}^{5} - 18x + 4 }$

$\displaystyle = \lim_{x \to \infty } \: \dfrac{ {x}^{5} \left( 3 + \dfrac{21}{ {x}^{2} } \right) }{ {x}^{5} \left( 7 - \dfrac{ 18}{ {x}^{4} } + \dfrac{ 4}{ {x}^{5} } \right) }$

$\displaystyle = \lim_{x \to \infty } \: \dfrac{ 3 + \dfrac{21}{ {x}^{2} } }{ 7 - \dfrac{ 18}{ {x}^{4} } + \dfrac{ 4}{ {x}^{5} } }$

$\displaystyle = \dfrac{ 3 + 0 }{ 7 - 0+ 0 }$

$\displaystyle = \dfrac{ 3 }{ 7 }$