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I'm taking my first calculus course and I'm having a bit of trouble solving this problem.

I've been trying to solve this for a while, but I don't even know where to start! I tried adding $U_{n}$ to the three members of the inequality above, but I don't think that it helps with anything.

There's a similar problem in my problem sheet, but I couldn't solve that one either so, maybe if I understand how to solve this one, I'll be able to solve the other one!

Could you guys help me out?

Thanks!

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  • $\begingroup$ Please include the actual question in the body of the post and not just the title, for future reference. $\endgroup$ – Nap D. Lover Oct 21 '18 at 20:57
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$$u_n^2>1\implies u_n>1$$

$$\implies \frac{1}{u_n}<1$$

$$\implies 1<u_n<\frac 1n +\frac{1}{u_n}$$

$$\implies 1<u_n<\frac 1n +1$$

$$\lim_{n\to+\infty} u_n=1$$

by squeeze theorem.

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  • $\begingroup$ Similar to one of the other answers here but simpler. +1 $\endgroup$ – Paramanand Singh Oct 22 '18 at 0:29
  • $\begingroup$ Oh, it's that simple! Thank you so much! :) $\endgroup$ – user94647 Oct 22 '18 at 8:53
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We have: $U_n^2 > 1\implies U_n > 1$ and $U_n <\dfrac{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^2}+4}}{2}=v_n$. We have: $1 \to 1$,and $v_n \to 1$ when $n \to \infty$. Thus by squeeze lemma $U_n \to 1$ as well.

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The map $$ f(x)=x-\frac{1}{x}$$ is strictly increasing and continuous on $(0,\infty)$. The inverse map $f^{-1}$is also continuous and $f^{-1}(0)=1$.

By hypothesis, the sequence $(f(U_n)) =(U_n-\frac{1}{U_n})$ converges to $0$. By continuity of $f^{-1}$, $U_n =f^{-1}(f(U_n))$ converges to $1$.

No need to compute square root or other stuff

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Have you proven the squeeze theorem yet? Notice that $U_n-\frac{1}{U_n}>0\implies U_n>\frac{1}{U_n}\implies U_n^2>1\implies U_n>1\implies1-\frac{1}{U_n}<\frac{1}{n}$. Also $0<\frac{1}{U_n}<1$, so $1-\frac{1}{U_n}>0$. Thus $$0<1-\frac{1}{U_n}<\frac{1}{n}\implies-1<\frac{-1}{U_n}<\frac{1}{n}-1.$$ By squeeze theorem $\frac{-1}{U_n}\to-1$. Apply algebraic limit properties to get $U_n\to1.$

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