# Continuity and limts: Show a function is continuous at a f(x)=(x+2x^3)^4,

Continuity and limts: Show a function is continuous at a
$f\left(x\right)={\left(x+2{x}^{3}\right)}^{4},a=-1$
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Show a function is continuous at a
$f\left(x\right)={\left(x+2{x}^{3}\right)}^{4},a=-1$
We have $f\left(x\right)={\left(x+2{x}^{3}\right)}^{4},a=-1$
$f\left(x\right)={\left(x+2{x}^{3}\right)}^{4}$ at a=-1 to be continous
if, L.H.L=R.H.L=Function at the given point
Now,
$\underset{x\to -1}{lim}f\left(x\right)=\underset{x\to -1}{lim}{\left(x+2{x}^{3}\right)}^{4}={\left(\underset{x\to -1}{lim}x+2{\left(\underset{x\to -1}{lim}x\right)}^{3}\right)}^{4}$
$⇒\underset{x\to -1}{lim}f\left(x\right)={\left(-1+2{\left(-1\right)}^{3}\right)}^{4}={\left(-1-2\right)}^{4}={\left(-3\right)}^{4}$
$⇒\underset{x\to -1}{lim}f\left(x\right)=81$
$\therefore L.H.L=\underset{x\to -1}{lim}f\left(x\right)=81$
Then,
$\underset{x\to 1}{lim}f\left(x\right)={\left(1+2{\left(1\right)}^{3}\right)}^{5}={\left(1+2\right)}^{4}={\left(3\right)}^{4}$
$\therefore R.H.L=\underset{x\to 1}{lim}f\left(x\right)=81$
Now, for f(-1)
$⇒f\left(-1\right)={\left(\left(-1\right)+2{\left(-1\right)}^{3}\right)}^{4}={\left(-1-2\right)}^{4}={\left(-3\right)}^{4}$
$⇒f\left(-1\right)=81$
Here, $L.H.L=R.H.L=f\left(-1\right)$
Hence, f(x) is continuous at a=-1