Limit proof problem as x goes to infinity

[bonfire09]

Homework Statement

prove that $\lim_{x \to \infty} \frac{\sqrt{x+1}}{x} = 0$ where $x>0$

Homework Equations

Definition: Let $A\subseteq\mathbb{R}$ and let $f:A\rightarrow \mathbb{R}$. Suppose that $(a,\infty)\subseteq A$ for some $a\in\mathbb{R}$. We say that $L\in\mathbb{R}$ is the limit of $f$ as $x\rightarrow\infty$ and write $\lim_{x \to \infty} f=L$ if any given $\epsilon>0$ there exists a $K=K(\epsilon)>a$ such that for any $x>K$ then $|f(x)-L|<\epsilon$.

The Attempt at a Solution

I can't seem to connect this definition to the problem because the part that is confusing is connecting $a$ to $\epsilon$. What I have so far is Suppose that $(1,\infty)\subseteq A$ where $a=1$. Let $\epsilon >0$ then there exists a $K\in\mathbb{N}$ such that if $x>K$ then... Let's just say I am completely stuck.

 

[tiny-tim]

hi bonfire09!

I can't seem to connect this definition to the problem because the part that is confusing is connecting $a$ to $\epsilon$.

you hae to prove that if x > a, then (√(x+1))/x < ε

 

[bonfire09]

Yeah but what do I let $\epsilon$ equal to? Do I let it be equal to $a$? Thats the part I am having trouble with. yes if $x>a$ then I can't really show $\frac{\sqrt{x+1}}{x}<\epsilon$.

 

[tiny-tim]

ahh! …

no, a is a function of ε ​

(in these problems, it's usually something like 1/ε or 1/√ε or 1/ε²)

 

[bonfire09]

Well it looks similar to the normal epsilon proofs for limits of sequences. Here is what I have for scratchwork. We want $|f(x)-0|<\epsilon \implies |\frac{\sqrt{x+1}}{x}|=\frac{\sqrt{x+1}}{x}\leq \sqrt{x+1}\leq x+1 <\epsilon$ for $x>0$. But then i am not sure what I should set my epsilon too?

Do I got about this way then? Let $\epsilon>1\implies \epsilon-1>0$ and let $a=\epsilon-1$ and there exists a $k>\epsilon-1$ such that if $x>k$... Would that be the way I should do it? THough I am not sure if that epsilon will work.

 

[tiny-tim]

Do I got about this way then? Let $\epsilon>1$

no, ε has to be very close to 0