Question

Evaluate :- $\lim_{x \to \infty} \left(\dfrac{\sqrt{x^2+1}-\sqrt[3]{x^3+1}}{\sqrt[4]{x^4+1}-\sqrt[5]{x^5+1}}\right)$

Answer 1

$\bf \to \: \green{{ \underline{answer}}} \\ \\ \sf \to \: \lim_{x \: \to \: \infty } \: = ( \frac{ \sqrt{ {x}^{2} + 1} - \sqrt[3]{ {x}^{3} + 1 } }{ \sqrt[4]{x {}^{4 } + 1 } - \sqrt[5]{x {}^{5} + 1 } } ) = 0$

$\bf \to \: \orange{{ \underline{step - \: by - \: step \: - explanation}}}$

$\sf \to \: \lim_{x \: \to \: \infty } \: = ( \dfrac{ \sqrt{x {}^{2} + 1} - \sqrt[3]{ {x}^{3} + 1 } }{ \sqrt[4]{ {x}^{4} + 1 } - \sqrt[5]{ {x}^{5} + 1} } ) \\ \\ \sf \to \: put \: the \: value \: of \: x \: = \infty \: \: in \: equation$

$\\ \\ \sf \to \: \lim_{x \: \to \: \infty } = ( \frac{ \sqrt{ \infty {}^{2} + 1} - \sqrt[3]{ \infty {}^{3} + 1} }{ \sqrt[4]{ \infty {}^{4} + 1 } - \sqrt[5]{ \infty {}^{5} + 1} } \\ \\ \sf \to \: it \: is \: in \: form \: of \: = \frac{ \infty }{ \infty }$

$\\ \\ \sf \to \: \lim_{x \to \: \infty } = ( \frac{ \sqrt{1 + \dfrac{1}{ {x}^{2} } } - \sqrt[3]{1 + \dfrac{1}{ {x}^{3} } } }{ \sqrt[4]{1 + \dfrac{1}{ {x}^{4} } } - \sqrt[5]{1 + \dfrac{1}{ {x}^{5} } } }$

$\\ \\ \sf \to \: \lim_{x \: \to \: \infty } = ( \frac{ \sqrt{1} - \sqrt{1} }{ \sqrt{1} - \sqrt{1} })$

$\\ \\ \sf \to \: \lim_{x \: \to \: \infty } = ( \frac{1 - 1}{1 - 1} ) \neq 0$

Not defined

Answer 2

$\tt To \: solve \\ \tt \lim_{x \to \infty} \left(\dfrac{\sqrt{x^2+1}-\sqrt[3]{x^3+1}}{\sqrt[4]{x^4+1}-\sqrt[5]{x^5+1}}\right) \\ \tt = \lim_{x \to \infty} \left(\dfrac{\sqrt{x^2+1}-\sqrt[3]{x^3+1}}{\sqrt[9]{x^4+1}} \right) \\ \tt = \lim_{x \to \infty} \frac{ \sqrt{ {x}^{2} (1 + \frac{1}{ {x}^{2} } ) - 3 \sqrt{ {x}^{3} } (1 + \frac{1}{ {x}^{3} } })}{ \sqrt [9]{ {x}^{4} + (1 + \frac{1}{ {x}^{2} } ) } } \\ \tt = \lim_{x \to \infty}( \frac{ \sqrt{1 + \frac{1}{x} } \sqrt [ - 3 \sqrt{3} ] {1 + \frac{1}{ {x}^{2} } } }{ \sqrt [ 9x ] {1 + \frac{1}{ \infin} } } \\ \tt = \frac{ \frac{1}{ \infin} \sqrt{1 + \frac{1}{ \infin} } \frac{ - 3}{ \sqrt{ \infin} } \sqrt{1 + \frac{1}{ \infin} } }{ \sqrt[ 9 ] {1 + \frac{1}{ \infin } }} \\ \tt = \frac{0 \sqrt{1 + 0} - 0 \sqrt{1 + 0} }{9 \sqrt{1 + 0} } \\ \tt = \frac{0 - 0}{9} = 0$