Question

If the limit given as $\underset{x\to \infty }{\mathop{\lim }}\,\left( \sqrt{\left( {{x}^{2}}-x+1 \right)}-ax-b \right)=0$, then for $k\ge 2$, $\underset{n\to \infty }{\mathop{\lim }}\,(se{{c}^{2n}}\left( k!\pi b \right))$ is equal to:
(a) –a
(b) a
(c) –b
(d) b

Answer 1

Hint: It is given the equation’s limit exists and is equal to zero. First convert the equation to indeterminate form, apply the limit and then equate it to zero, you will get one variable value. After deriving numerator and denominator separately and then applying the limit you will get the other variable value. Using this value evaluates the required limit.

Complete step-by-step answer:
It is given that the limit of a function exists and is equal to zero, i.e.,
$\underset{x\to \infty }{\mathop{\lim }}\,\left( \sqrt{\left( {{x}^{2}}-x+1 \right)}-ax-b \right)=0$
Applying the limits we get LHS as infinity which is not equal to zero.
And we know L’Hospital’s Rule won’t work on products, it only works on quotients.
So now we will convert the above equation into quotient form.
Let $x=\dfrac{1}{h}$ .
Now
$\begin{align}
  & x\to \infty \\
 & \Rightarrow h\to 0 \\
\end{align}$
So the given limit can be written as,
$\underset{h\to 0}{\mathop{\lim }}\,\left( \sqrt{\left( {{\left( \dfrac{1}{h} \right)}^{2}}-\dfrac{1}{h}+1 \right)}-a\left( \dfrac{1}{h} \right)-b \right)=0$
Taking the LCM under the root, we get
$\begin{align}
  & \underset{h\to 0}{\mathop{\lim }}\,\left( \sqrt{\left( \dfrac{1-h+{{h}^{2}}}{{{h}^{2}}} \right)}-a\left( \dfrac{1}{h} \right)-b \right)=0 \\
 & \Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\sqrt{\left( 1-h+{{h}^{2}} \right)}}{h}-a\left( \dfrac{1}{h} \right)-b \right)=0 \\
\end{align}$
Now again taking the LCM, we get
$\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\sqrt{\left( 1-h+{{h}^{2}} \right)}-a-bh}{h} \right)=0.......(i)$
Applying the limit, we get
$\begin{align}
  & \Rightarrow \left( \dfrac{\sqrt{\left( 1-0+{{0}^{2}} \right)}-a-b(0)}{(0)} \right)=0 \\
 & \Rightarrow \sqrt{1}-a-0=0 \\
 & \Rightarrow a=1..........(i) \\
\end{align}$
It is given that the limit of a function exists and is equal to zero, i.e.,
$\underset{x\to \infty }{\mathop{\lim }}\,\left( \sqrt{\left( {{x}^{2}}-x+1 \right)}-ax-b \right)=0$
And we have converted this into quotient form as in equation (i) and we can now use L’Hospital’s Rule, i.e., differentiate numerator and denominator separately before applying the limits.
Consider equation (i):
$\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\sqrt{\left( 1-h+{{h}^{2}} \right)}-a-bh}{h} \right)=0$
Differentiating numerator and denominator separately, we get
$\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\dfrac{d}{dh}\left( \sqrt{\left( 1-h+{{h}^{2}} \right)}-a-bh \right)}{\dfrac{d}{dh}\left( h \right)} \right)=0$
Applying the sum rule of differentiation in numerator, we get
$\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\dfrac{d}{dh}\left( \sqrt{\left( 1-h+{{h}^{2}} \right)} \right)-\dfrac{d}{dh}\left( a \right)-\dfrac{d}{dh}\left( bh \right)}{1} \right)=0$
Now differentiation of constant term is zero, so above equation becomes,
\[\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\dfrac{1}{2\sqrt{\left( 1-h+{{h}^{2}} \right)}}\dfrac{d}{dh}\left( 1-h+{{h}^{2}} \right)-0-b\dfrac{d}{dh}\left( h \right)}{1} \right)=0\]
Again applying the sum rule of differentiation in numerator, we get
\[\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\dfrac{1}{2\sqrt{\left( 1-h+{{h}^{2}} \right)}}\left( \dfrac{d}{dh}\left( 1 \right)-\dfrac{d}{dh}\left( h \right)+\dfrac{d}{dh}\left( {{h}^{2}} \right) \right)-b\left( 1 \right)}{1} \right)=0\]
Now differentiation of constant term is zero, so above equation becomes,
\[\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\dfrac{1}{2\sqrt{\left( 1-h+{{h}^{2}} \right)}}\left( 0-\left( 1 \right)+2h \right)-b}{1} \right)=0\]
Solving this, we get
\[\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{-1+2h}{2\sqrt{\left( 1-h+{{h}^{2}} \right)}}-b \right)=0\]
Applying the limits, we get
\[\Rightarrow \left( \dfrac{-1+2(0)}{2\sqrt{\left( 1-(0)+{{(0)}^{2}} \right)}}-b \right)=0\]
Solving this, we get
\[\begin{align}
  & \Rightarrow \left( \dfrac{-1}{2\sqrt{1}}-b \right)=0 \\
 & \Rightarrow b=-\dfrac{1}{2}........(ii) \\
\end{align}\]
Now we got the values of ‘a’ and ‘b’.
Consider the function whose limit needs to be found, i.e.,
$\underset{n\to \infty }{\mathop{\lim }}\,(se{{c}^{2n}}\left( k!\pi b \right))..........(iii)$
It is given that $k\ge 2$, so k factorials can be written as,
$\begin{align}
  & 2!=2\times 1 \\
 & 3!=3\times 2\times 1 \\
 & 4!=4\times 3\times 2\times 1 \\
\end{align}$
We can see that the factorial has repeated pattern, so general term of k! when $k\ge 2$ can be written as
$k!=N\times 2$ where N = 1, 2, 3, ….
Substituting this value and value of ‘b’ in equation (iii), we get
$\underset{n\to \infty }{\mathop{\lim }}\,(se{{c}^{2n}}\left( k!\pi b \right))=\underset{n\to \infty }{\mathop{\lim }}\,\left( se{{c}^{2n}}\left( 2N\pi \left( -\dfrac{1}{2} \right) \right) \right)$
Cancelling the like terms, we get
$\underset{n\to \infty }{\mathop{\lim }}\,(se{{c}^{2n}}\left( k!\pi b \right))=\underset{n\to \infty }{\mathop{\lim }}\,\left( se{{c}^{2n}}\left( -N\pi \right) \right)$
We also know, $\sec \left( -\theta \right)=\sec \theta $, so above equation can be written as,
$\underset{n\to \infty }{\mathop{\lim }}\,(se{{c}^{2n}}\left( k!\pi b \right))=\underset{n\to \infty }{\mathop{\lim }}\,{{\left( sec\left( N\pi \right) \right)}^{2n}}$
Now we know $\sec (n\pi )=-1$, so the above equation can be written as,
\[\begin{align}
  & \underset{n\to \infty }{\mathop{\lim }}\,(se{{c}^{2n}}\left( k!\pi b \right))=\underset{n\to \infty }{\mathop{\lim }}\,{{\left( -1 \right)}^{2n}} \\
 & \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,(se{{c}^{2n}}\left( k!\pi b \right))=\underset{n\to \infty }{\mathop{\lim }}\,{{\left( 1 \right)}^{n}} \\
\end{align}\]
Now one raised to the infinity is always one, so the answer is
$\underset{n\to \infty }{\mathop{\lim }}\,(se{{c}^{2n}}\left( k!\pi b \right))=1$
Substituting value from equation (i), we get
$\underset{n\to \infty }{\mathop{\lim }}\,(se{{c}^{2n}}\left( k!\pi b \right))=a$
Hence the correct answer is option (b).

Note: The possibility of mistake is after seeing the question the student apply the limits directly,
$\underset{x\to \infty }{\mathop{\lim }}\,\left( \sqrt{\left( {{x}^{2}}-x+1 \right)}-ax-b \right)=0$ we get $\infty \ne 0$.
So the answer will be wrong.
So we should convert the given equation into quotient form.
Students will get confused to find the ‘k’ factorial also.