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<14x&text{if}&xgeq1end{cases} lim_{xrightarrow1} f(x)

Question
Limits and continuity
asked 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. \(f(x)=\begin{cases}\sqrt[3]{x^2+7}&\text{if}&x<1\\4x&\text{if}&x\geq1\end{cases}\)</span>
\(\lim_{x\rightarrow1} f(x)\)

Answers (1)

2021-02-04
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.
0

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