Calculate: $$\lim_{x \to +\infty}\frac{x+\sqrt{x}}{x-\sqrt{x}}$$ i tried to take $X=\sqrt{x}$ we give us
when $x \to 0$ we have $X \to 0$ But i really don't know if it's a good idea
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I suggest you take your analysis textbook and try to understand limits a little better before you attempt an exercise. – AnalysisStudent0414 Feb 25 '16 at 23:12
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@AnalysisStudent0414 i find that it's 1 is that true ? – user315918 Feb 25 '16 at 23:13
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@user315918 It is! You may want to check the answers down below for some hints on how to solve it if you'd like to see other methods. Alternatively, you could post your own so others can review your work. – zz20s Feb 25 '16 at 23:22
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Yes, it is true. However I really encourage you to look more at the "What am I doing?" and less at "How is it done?", so you can understand what is a good idea and what is not. – AnalysisStudent0414 Feb 26 '16 at 00:14
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You may have noticed the similarity between the nominator and the denominator. If you choose to make use of this you can express the ratio in the following way. $$\frac{x+\sqrt{x}}{x-\sqrt{x}}=\frac{(x-\sqrt{x})+2\sqrt{x}}{x-\sqrt{x}} = 1+\frac{2}{\sqrt{x}-1}\ .$$ If $x$ is a large positive number, then so will its square-root be a large positive number; consequently the original ratio will be close to $1$. If you like, you can formalize this using an $\epsilon$-$\delta$-argument, but I will leave that to you.
Anders Muszta
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divide through by $x$ to get
$$ \frac{ 1 + 1/\sqrt{x}}{1-1/\sqrt{x}} $$ It is now rather obvious.
Mark Joshi
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