I'm having trouble with these limits. Any help would be appreciated. $\displaystyle \lim_{x\to +\infty}\frac{\sqrt[3]{{x}^{5}-3x}}{{x}^{2}+4}$ $\displaystyle \lim_{x\to 0^{+}} \sqrt{\frac{x+1}{x}}$
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Originally Posted by Alvy I'm having trouble with these limits. Any help would be appreciated. $\displaystyle \lim_{x\to +\infty}\frac{\sqrt[3]{{x}^{5}-3x}}{{x}^{2}+4}$ $\displaystyle \lim_{x\to 0^{+}} \sqrt{\frac{x+1}{x}}$ Observe: $\displaystyle \frac{\sqrt[3]{{x}^{5}-3x}}{{x}^{2}+4}=\frac{x^{5/3}(1-3/x^4)}{x^2(1+4/x^2)}=\frac{(1-3/x^4)}{x^{1/3}(1+4/x^2)}$ Now you should be able to determine the limit. CB
Originally Posted by Alvy I'm having trouble with these limits. Any help would be appreciated. $\displaystyle \lim_{x\to +\infty}\frac{\sqrt[3]{{x}^{5}-3x}}{{x}^{2}+4}$ $\displaystyle \lim_{x\to 0^{+}} \sqrt{\frac{x+1}{x}}$ For the second one: $\displaystyle \sqrt{\frac{x+1}{x}} = \sqrt{\frac{\frac{x+1}{x}}{\frac{x}{x}}} = \sqrt{\frac{1+{\frac{1}{x}}}{1}}= \sqrt {1+{\frac{1}{x}}}$
Thank you very much.
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