A piecewise function is given. Use properties of limits to find the shown indicated limit, or

Given
$f(x)=\{\begin{array}{lll}\sqrt[3]{x^2+7}& \text{if}& x<1\\ 4x& \text{if}& x\ge 1\end{array}$
we know for any given function g(x) limit at given point x=h exist only
if, $\underset{x->h^{-}}{lim}g(x)=\underset{x->h^{+}}{lim}g(x),$
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,
$L.H.L=\underset{x\to 1^{-}}{lim}f(x)$
$(\text{if }x<1,\text{then}f(x)=\sqrt[3]{x^2+7}$
$=\underset{x\to 1^{-}}{lim}\sqrt[3]{x^2+7}$
$=\sqrt[3]{1^2+7}$
$=\sqrt[3]{1+7}$
$=\sqrt[3]{8}$
$=2$
Now finding right hand limit also
$R.H.L=\underset{x\to 1^{+}}{lim}f(x)$
$(\text{if }x>1\text{ then }f(x)=4x)$
$=\underset{x\to 1^{+}}{lim}(4x)$
$=\underset{x{\to }^{+}}{lim}4(1)$
$=4$
here
L.H.L != R.H.L.
hence, for given function limit at x=1 doesn't exists.