Question

Solve this limit question.

Answer 1

$lt \: \: x = > \infty \\ \\ lt \: \frac{1}{x} = > 0 \\ \\ now \\ \\ lt \: \: x = > \infty \: \: \: \: \frac{x}{1 + {x}^{2} + {x}^{4} } \\ \\ divide \: with \: {x}^{4} \\ \\ lt \: \: x = > \infty \: \: \: \frac{ \frac{1}{ {x}^{3} } }{ \frac{1}{ {x}^{4} } + \frac{1}{ {x}^{2} } + 1} \\ \\ = \frac{0}{0 + 0 + 1} \\ \\ = 0$

hope it helps

Answer 2

Answer:

0

Step-by-step explanation:

$Given:\frac{r}{1+r^2 + r^4}$

Divide the numerator and denominator by r⁴

$\Longrightarrow \frac{\frac{r}{r^4}}{\frac{1}{r^4}+\frac{r^2}{r^4}+\frac{r^4}{r^4}}$

$\Longrightarrow \frac{\frac{1}{r^3}}{\frac{1}{r^4}+\frac{1}{r^2}+1}$

$\Longrightarrow\frac{ \lim_{r\to\infty} (\frac{1}{r^3})}{\lim_{r\to\infty} (\frac{1}{r^4}+\frac{1}{r^2} + 1)}$

(i)

$\Longrightarrow\lim_{r\to\infty} \frac{1}{r^3}$

$\Longrightarrow0$

(ii)

$\Longrightarrow\lim_{r \to \infty} (\frac{1}{r^4} + \frac{1}{r^2} + 1)$

$\Longrightarrow 0 + 0 + 1$

$\Longrightarrow 1$

On simplifying, we get

$\Longrightarrow \frac{0}{1}$

$\Longrightarrow\boxed{0}$

Hope it helps!