Prove that lim of x/(x+1) = 1 as x approaches infinity

Question

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.

Answer 1

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 $1?$

Answer 2

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 $$ x+1>\frac{1}{\varepsilon}. $$

Answer 3

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 $\lim_{x \to \infty} \frac1x = 0$

Answer 4

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 $\displaystyle \lim_{x \to \infty} \left(\dfrac{x}{x+1} -1\right) =0 \Rightarrow \displaystyle \lim_{x\to \infty} \dfrac{x}{x+1} = 1$.

Answer 5

I believe a much simpler approach is to write $x$ as $x + 1 - 1$ and then you get $\lim_{x\rightarrow \infty } \frac{x + 1 - 1}{x + 1} = \lim_{x\rightarrow \infty }\left ( 1 - \frac{1}{x+1} \right )$

p.s. this kinda "trick" can be useful when integrating too ;)