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

nitraiddQ 2021-02-03 Answered

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

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

Maciej Morrow
Answered 2021-02-04 Author has 4713 answers

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.

Have a similar question?
Ask An Expert
25
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more
...