Question

The given function is \[F(x)=\left( \begin{align}
  & \dfrac{x+3{{x}^{2}}+5{{x}^{3}}+...+(2n-1){{x}^{n}}-{{n}^{2}}}{x-1};x\ne 1 \\
 & \dfrac{n({{n}^{2}}-1)}{3};x=1 \\
\end{align} \right)\] . Check if the function is continuous at x = 1

Answer 1

Hint: Now we know that the condition for the function to be continuous is $\underset{x\to 1}{\mathop{\lim }}\,F(x)=F(1)$ .
Hence first we will find $\underset{x\to 1}{\mathop{\lim }}\,F(x)$ . To remove indeterminate form L hospital rule. Now that the indeterminate form is removed we can substitute x = 1 to find $\underset{x\to 1}{\mathop{\lim }}\,F(x)$. Now we know that $\sum{n}=\dfrac{n\left( n+1 \right)}{2}$ and $\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ . we will use this to simplify the expression further. Now we know that \[F(1)=\dfrac{n({{n}^{2}}-1)}{3}\] Hence, we will check if $\underset{x\to 1}{\mathop{\lim }}\,F(x)=F(1)$ and determine if the function is continuous.

Complete step by step answer:
Now if function is continuous at x = 1 we have $\underset{x\to 1}{\mathop{\lim }}\,F(x)=F(1)$
We know that \[F(1)=\dfrac{n({{n}^{2}}-1)}{3}...................................(1)\]
Now consider $L=\underset{x\to 1}{\mathop{\lim }}\,F(x)$.
Hence we have \[L=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{x+3{{x}^{2}}+5{{x}^{3}}+...+(2n-1){{x}^{n}}-{{n}^{2}}}{x-1}\]
Now to solve this limit we will use the L-Hospital rule.
According to L-Hospital rule we have $\underset{x\to 1}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{f'(x)}{g'(x)}$
Now we know that $\dfrac{d(f+g)}{dx}=\dfrac{df}{dx}+\dfrac{dg}{dx}$ and for any non-zero n we have $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$
Hence we get
$\begin{align}
  & L=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{1+6x+15{{x}^{2}}+...+n(2n-1){{x}^{n-1}}-0}{-1} \\
 & \Rightarrow L=\underset{x\to 1}{\mathop{\lim }}\,\left( 1+6x+15{{x}^{2}}+...+n(2n-1){{x}^{n-1}}-0 \right) \\
\end{align}$
Now we can substitute x = 1, to solve the limit.
\[L=\left( 1+6+15+...+n(2n-1) \right)\]
Now this is nothing but
$\begin{align}
  & L=\left( 1+2\left( 2\left( 2 \right)-1 \right)+3\left( 2\left( 3 \right)-1 \right) \right)+...n(2n-1) \\
 & L=\left( \sum{n(2n-1)} \right) \\
\end{align}$
Now multiplying the terms n and (2n – 1) we get
$L=\sum{\left( 2{{n}^{2}}-n \right)}$
Now we know that $\sum{\left( a-b \right)}=\sum{a}-\sum{b}$
 Hence we have
$L=\left( \sum{\left( 2{{n}^{2}} \right)}-\sum{n} \right)................\left( 2 \right)$
Now we have $\sum{n}=\dfrac{n\left( n+1 \right)}{2}$ and $\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
Now we will substitute this expansion in (2).
$L=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{3}-\dfrac{n(n+1)}{2}$
Taking LCM we get
$\begin{align}
  & L=\dfrac{2n\left( n+1 \right)\left( 2n+1 \right)-3n\left( n+1 \right)}{6} \\
 & \Rightarrow L=\dfrac{n\left( n+1 \right)[2\left( 2n+1 \right)-3]}{6} \\
 & \Rightarrow L=\dfrac{n\left( n+1 \right)[4n+2-3]}{6} \\
 & \Rightarrow L=\dfrac{n\left( n+1 \right)\left( 4n-1 \right)}{6} \\
 & \therefore L=\underset{x\to 1}{\mathop{\lim }}\,F(x)=\dfrac{n\left( n+1 \right)\left( 4n-1 \right)}{6} \\
\end{align}$
Hence we get $\underset{x\to 1}{\mathop{\lim }}\,F(x)=\dfrac{n\left( n+1 \right)\left( 4n-1 \right)}{6}............................\left( 3 \right)$
Now from equation (1) and equation (3) we get.
$\underset{x\to 1}{\mathop{\lim }}\,F(x)\ne F(1)$
Hence the function is not continuous at x = 1.

Note:
Now here we can also solve this limit by dividing \[x+3{{x}^{2}}+5{{x}^{3}}+...+(2n-1){{x}^{n}}-{{n}^{2}}\] by (x – 1), then we will have the limit in solvable form as (x – 1) from denominator will be cancelled.
Also we can solve this in short cut method
Once we have $L=\underset{x\to 1}{\mathop{\lim }}\,\left( 1+6x+15{{x}^{2}}+...+n(2n-1){{x}^{n-1}}-0 \right)$ we can check the value for n = 2 and x = 1 which will be 1 + 6 = 7
On the other hand for n = 2 we have \[\dfrac{n({{n}^{2}}-1)}{3}\] = 2.
Hence we have both values are not equal and the function is discontinuous