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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