Given

\(f(x)=\begin{cases}\sqrt[3]{x^2+7}&\text{if}&x<1\\4x&\text{if}&x\geq1\end{cases}\)</span>

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,

\(L.H.L=\lim_{x\rightarrow1^-} f(x)\)

\((\text{if }x<1,\text{then}f(x)=\sqrt[3]{x^2+7}\)</span>

\(=\lim_{x\rightarrow1^-}\sqrt[3]{x^2+7}\)

\(=\sqrt[3]{1^2+7}\)

\(=\sqrt[3]{1+7}\)

\(=\sqrt[3]{8}\)

\(=2\)

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.

\(f(x)=\begin{cases}\sqrt[3]{x^2+7}&\text{if}&x<1\\4x&\text{if}&x\geq1\end{cases}\)</span>

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,

\(L.H.L=\lim_{x\rightarrow1^-} f(x)\)

\((\text{if }x<1,\text{then}f(x)=\sqrt[3]{x^2+7}\)</span>

\(=\lim_{x\rightarrow1^-}\sqrt[3]{x^2+7}\)

\(=\sqrt[3]{1^2+7}\)

\(=\sqrt[3]{1+7}\)

\(=\sqrt[3]{8}\)

\(=2\)

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.