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Question Number 144564 by imjagoll last updated on 26/Jun/21

Commented by justtry last updated on 26/Jun/21

$$0??$$

Answered by liberty last updated on 26/Jun/21

$$\underset{x\to -\mathrm{\infty }}{\lim }\left(\frac{{(\mathrm{x}-\sqrt{{\mathrm{x}}^2-2\mathrm{x}+1})}^{2021}}{{\mathrm{x}}^{2021}}\right)+\underset{x\to -\mathrm{\infty }}{\lim }\left(\frac{{(\mathrm{x}-\sqrt{{\mathrm{x}}^2+5})}^{2021}}{{\mathrm{x}}^{2021}}\right)$$$$=\underset{x\to -\mathrm{\infty }}{\lim }{\left(\frac{\mathrm{x}+\mathrm{x}\sqrt{1-{2\mathrm{x}}^{-1}+{\mathrm{x}}^{-2}}}{\mathrm{x}}\right)}^{2021}+\underset{x\to -\mathrm{\infty }}{\lim }{\left(\frac{\mathrm{x}+\mathrm{x}\sqrt{1+{5\mathrm{x}}^{-2}}}{\mathrm{x}}\right)}^{2021}$$$$=2^{2021}+2^{2021}=2^{2022}$$

Commented by justtry last updated on 26/Jun/21

$$nice,thankyou$$

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