# Prove $\lim_{n\to\infty}U_{n} = 1$ given $0 \lt U_{n} - {1\over U_{n}}\lt {1\over n}$ and $U_n>0$

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!

• Please include the actual question in the body of the post and not just the title, for future reference. – Nap D. Lover Oct 21 '18 at 20:57

$$u_n^2>1\implies u_n>1$$

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

$$\implies 1

$$\implies 1

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

by squeeze theorem.

• Similar to one of the other answers here but simpler. +1 – Paramanand Singh Oct 22 '18 at 0:29
• Oh, it's that simple! Thank you so much! :) – user94647 Oct 22 '18 at 8:53

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.

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

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 $$U_n\to1.$$