 nitraiddQ

2021-02-03

A piecewise function is given. Use properties of limits to find the shown indicated limit, or state that the limit does not exist. $f\left(x\right)=\left\{\begin{array}{lll}\sqrt{{x}^{2}+7}& \text{if}& x<1\\ 4x& \text{if}& x\ge 1\end{array}$
$\underset{x\to 1}{lim}f\left(x\right)$ Maciej Morrow

Given
$f\left(x\right)=\left\{\begin{array}{lll}\sqrt{{x}^{2}+7}& \text{if}& x<1\\ 4x& \text{if}& x\ge 1\end{array}$
we know for any given function g(x) limit at given point x=h exist only
if, $\underset{x->{h}^{-}}{lim}g\left(x\right)=\underset{x->{h}^{+}}{lim}g\left(x\right),$
thus, we are finding left hand limit and right hand limit one by one at x=1, if both are coming same then limit will exist other limit of given function will not exist
so,
$L.H.L=\underset{x\to {1}^{-}}{lim}f\left(x\right)$

$=\underset{x\to {1}^{-}}{lim}\sqrt{{x}^{2}+7}$
$=\sqrt{{1}^{2}+7}$
$=\sqrt{1+7}$
$=\sqrt{8}$
$=2$
Now finding right hand limit also
$R.H.L=\underset{x\to {1}^{+}}{lim}f\left(x\right)$

$=\underset{x\to {1}^{+}}{lim}\left(4x\right)$
$=\underset{x{\to }^{+}}{lim}4\left(1\right)$
$=4$
here
L.H.L != R.H.L.
hence, for given function limit at x=1 doesn't exists.

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