Anonym

2021-10-26

Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.
$f\left(x\right)={\left(x+2{x}^{3}\right)}^{4},a=-1$

falhiblesw

Limits
Limit of the function at point a exists if and only if left hand limit and right hand limit are equal and exists finitely at point a.
Left hand limit
$\underset{x\to {a}^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f\left(a-h\right)$
Right hand limit
$\underset{x\to {a}^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f\left(a+h\right)$
So, limit exists at pont a if and only if $\underset{x\to {a}^{-}}{lim}f\left(x\right)=\underset{x\to {a}^{+}}{lim}f\left(x\right)=$ finite
Continuity
A function is said to be continuous at a if and only if
1. Limit exits that is $\underset{x\to a}{lim}f\left(x\right)=\underset{x\to {a}^{+}}{lim}f\left(x\right)=$ finite
2. Limit is equal to value of the function at point a that is $\underset{x\to {a}^{-}}{lim}f\left(x\right)=\underset{x\to {a}^{+}}{lim}f\left(x\right)=f\left(a\right)$
Solution:
We have given function $f\left(x\right)={\left(x+2{x}^{3}\right)}^{4}$ at points a=-1
Left hand limit
$\underset{x\to {a}^{+}}{lim}f\left(x\right)=\underset{x\to -{1}^{+}}{lim}{\left(x+2{x}^{3}\right)}^{4}=\underset{h\to 0}{lim}{\left(\left(-1+h\right)+2{\left(-1+h\right)}^{3}\right)}^{4}$
$={\left(-1+2{\left(-1\right)}^{3}\right)}^{4}$
$={\left(-1-2\right)}^{4}={\left(-3\right)}^{4}=81$
Hence $\underset{x\to -{1}^{+}}{lim}f\left(x\right)=81$
So limit exists
Now, $f\left(-1\right)={\left(-1+2{\left(-1\right)}^{3}\right)}^{4}$
$={\left(-1-2\right)}^{4}={\left(-3\right)}^{4}=81$
Therefore,
Given function is continuous at x=-1

Jeffrey Jordon