Hey please i need some small tips to get me starting. I tried multiplying with the conjugate but it didnt bring me very far..
How to find $\lim\limits_{n\to \infty} \frac{\frac{n}{1000}}{\sqrt{n+\frac{n}{1000}}+\sqrt{n}}$
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real-analysis
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0Lol it doesn't even converge – 2017-01-08
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0What does your question mean? – 2017-01-08
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0@mathJuan He wants to know how to evaluate the infinite limit. – 2017-01-08
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0im sorry i wanted to know why the limit goes to infinity – 2017-01-08
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0The limit doesn't go anywhere. The terms go somewhere though. If the problem is to show the limit is $\infty,$ why didn't you say so in your question? – 2017-01-08
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0$$\frac{\frac{n}{1000}}{\sqrt{n+\frac{n}{1000}}+\sqrt{n}}=\frac{n\frac{1}{1000}}{\sqrt{n}\sqrt{1+\frac{1}{1000}}+\sqrt{n}}=\frac{n}{\sqrt{n}}\frac{\frac{1}{1000}}{\sqrt{1+\frac{1}{1000}}+\sqrt{1}}=c\sqrt{n}$$ – 2017-01-08
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0@ Did: Yes, I wait a solution for your last question. – 2017-01-08
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0@mathJuan Which last question? (Unrelated: No space between @ and username, please.) – 2017-01-08
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0@Did Thank you. I want to say about the question "why didn't you say so in your question?" . – 2017-01-08
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0@mathJuan User zhw. asked this, not I. – 2017-01-08
2 Answers
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To compute the limit, note that $\sqrt{n + \frac{n}{1000}} + \sqrt{n} \leq \sqrt{4n} + \sqrt{n} = 3\sqrt{n}$. Thus $$\frac{\frac{n}{1000}} {\sqrt{n + \frac{n}{1000}} + \sqrt{n}} \geq \frac{\frac{n}{1000}}{3\sqrt{n}} = \frac{1}{3000}\sqrt{n}$$ which goes to $\infty$ as $n \to \infty$.
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0excuse me where did the n at the end go ? – 2017-01-08
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1$n/\sqrt{n} = \sqrt{n}$. – 2017-01-08
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0oh yeah ahahh thanks – 2017-01-08
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0shouldnt it be sqrt(n)/3000 ? – 2017-01-08
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0Ahh yes, you're correct. However, the constant does not change the limit behavior of the function. – 2017-01-08
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Set $c=\sqrt{1001/1000}$; then you have $$ \lim_{n\to\infty}\frac{1}{1000(c+1)}\frac{n}{\sqrt{n}}= \lim_{n\to\infty}\frac{\sqrt{n}}{1000(c+1)}=\infty $$ This is because $$ \sqrt{n+\frac{n}{1000}}=\sqrt{\frac{1001}{1000}n}=c\sqrt{n} $$ Thus you have $$ \frac{\dfrac{n}{1000}}{c\sqrt{n}+\sqrt{n}}= \frac{1}{1000(c+1)}\frac{n}{\sqrt{n}} $$