Findf(x), where f(x) =

Question:

Find $\lim _{x \rightarrow 5} 1(x)$, where $t(x)=|x|-5$

Solution:

The given function is $f(x)=|x|-5$.

$\lim _{x \rightarrow 5} f(x)=\lim _{x \rightarrow 5^{-}}[|x|-5]$

$=\lim _{x \rightarrow 5}(x-5) \quad[$ When $x>0,|x|=x]$

$=5-5$

$=0$

$\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5^{+}}(|x|-5)$

$=\lim _{x \rightarrow 5}(x-5) \quad[$ When $x>0,|x|=x]$

$=5-5$

$=0$

$\therefore \lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5^{+}} f(x)=0$

Hence, $\lim _{x \rightarrow 5} f(x)=0$

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