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

A piecewise function is given. Use properties of limits to find the shown indicated limit, or state that the limit does not exist. $$f(x)=\begin{cases}\sqrt[3]{x^2+7}&\text{if}&x<1\\4x&\text{if}&x\geq1\end{cases}$$
$$\lim_{x\rightarrow1} f(x)$$

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Maciej Morrow

Given
$$f(x)=\begin{cases}\sqrt[3]{x^2+7}&\text{if}&x<1\\4x&\text{if}&x\geq1\end{cases}$$
we know for any given function g(x) limit at given point x=h exist only
if, $$\lim_{x->h^-}g(x)=\lim_{x->h^+}g(x),$$
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,
$$L.H.L=\lim_{x\rightarrow1^-} f(x)$$
$$(\text{if }x<1,\text{then}f(x)=\sqrt[3]{x^2+7}$$
$$=\lim_{x\rightarrow1^-}\sqrt[3]{x^2+7}$$
$$=\sqrt[3]{1^2+7}$$
$$=\sqrt[3]{1+7}$$
$$=\sqrt[3]{8}$$
$$=2$$
Now finding right hand limit also
$$R.H.L=\lim_{x\rightarrow1^+}f(x)$$
$$(\text{if } x>1\text{ then } f(x)=4x)$$
$$=\lim_{x\rightarrow1^+}(4x)$$
$$=\lim_{x\rightarrow^+}4(1)$$
$$=4$$
here
L.H.L != R.H.L.
hence, for given function limit at x=1 doesn't exists.