# How does \lim_(n\rarr\infty)(2|x-2|)/(n+3)=0?

## Without doing any further work, is it possible to recognize the limit as 0? Or should I simplify further (if so, then how?)

Apr 26, 2018

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#### Explanation:

show the steps and focus in the explanition

since $n \rightarrow \infty$ n is a variable value and others like: x are constant

x=constant

any constant divide by$\infty$equal zero

c=any constant
$\frac{c}{\infty} = 0$

$\frac{c}{-} \infty = 0$

in addition any constant plus $\infty$ or $- \infty$ equal
$c + \left(+ \infty\right) = \infty$ and
$c + \left(- \infty\right) = - \infty$

${\lim}_{n \rightarrow \infty} \left[\frac{2 | x - 2 |}{n + 3}\right] = \frac{2 | x - 2 |}{\infty} = 0$

$\left(2 | x - 2 |\right) =$constant

${\lim}_{n \rightarrow \infty} \left[\frac{2 | x - 2 |}{n + 3}\right] = 0$