Prove that lim of x/(x+1) = 1 as x approaches infinity
Andrew Henderson 
I want to prove that $$\lim\limits_{x\to \infty} \frac{x}{x+1}=1$$
I know that I need to show that: $$\left|\frac{1}{x+1}\right| \lt \epsilon$$ But I'm not sure how to manipulate it. Any help or hint would be appreciated.
$\endgroup$ 14 Answers
$\begingroup$Rearrange it as follows: $$\lim_{x\to\infty}\dfrac{x}{x+1}=\lim_{x\to\infty}\dfrac{1}{\dfrac{1}{x}+1}$$ Can you see why the limit is $1?$
$\endgroup$ $\begingroup$If $x>-1$, then $x+1>0$, so you can drop the absolute values $$ \frac{1}{x+1}<\varepsilon. $$ Note that this is equivalent to $$ x+1>\frac{1}{\varepsilon}. $$
$\endgroup$ 3 $\begingroup$Why so complicated over $\varepsilon - \delta $ ? I suggest $ \frac{x}{x+1} = \frac{1}{1+1/x} $, thus
$\lim_{x \to \infty} \frac{1}{1+1/x} = 1 $ since $\lim_{x \to \infty} \frac1x = 0$
$\endgroup$ $\begingroup$We have: $0 < \left|\dfrac{x}{x+1} - 1\right| = \dfrac{1}{|x+1|} < \dfrac{1}{|x|} = \dfrac{1}{x}$. Use squeeze lemma to get $\displaystyle \lim_{x \to \infty} \left(\dfrac{x}{x+1} -1\right) =0 \Rightarrow \displaystyle \lim_{x\to \infty} \dfrac{x}{x+1} = 1$.
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