Question

If the function \[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}}}&{;x \ne \pi } \\
  k&{;x = \pi }
\end{array}} \right.\] is continuous at\[x = \pi \], then k equals:
A) 0
B) \[\dfrac{1}{2}\]
C) 2
D) \[\dfrac{1}{4}\]

Answer 1

Hint: Here first we will evaluate the limit at \[x = \pi \] of the function f(x) when \[x \ne \pi \] and then equate it with k to get its value as the given function is continuous.
L’ hospital rule states that if on evaluating the limit of a function we get an indeterminate form i.e. \[\dfrac{0}{0} or \dfrac{\infty }{\infty }\] then we need to differentiate the numerator and the denominator of the function separately and then apply the limit to get the answer.

Complete step-by-step answer:
The given function is:-
\[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}}}&{;x \ne \pi } \\
  k&{;x = \pi }
\end{array}} \right.\]
Now we will first evaluate the limit of f(x) at \[x = \pi \]using the value of function when \[x \ne \pi \]
Hence on writing the limit we get:-
\[\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}}\]
Now on putting in the value of limit we get:-
\[ = \dfrac{{\sqrt {2 + cos\pi } - 1}}{{{{(\pi - \pi )}^2}}}\]
Now we know that:
\[cos\pi = - 1\]
Hence putting the value we get:-
\[ = \dfrac{{\sqrt {2 - 1} - 1}}{{{{(\pi - \pi )}^2}}}\]
Solving it further we get:-
\[
   = \dfrac{{\sqrt 1 - 1}}{{{{(\pi - \pi )}^2}}} \\
   = \dfrac{0}{0} \\
 \]
Now since we got an indeterminate form i.e. \[\dfrac{0}{0}\]
Hence we have to apply L’ hospital rule to evaluate the limit.
Now, L’ hospital rule states that if on evaluating the limit of a function we get an indeterminate form i.e. \[\dfrac{0}{0}or\dfrac{\infty }{\infty }\] then we need to differentiate the numerator and the denominator of the function separately and then apply the limit to get the answer.
Hence we will differentiate the numerator and the denominator of the given function separately
Hence on differentiating we get:-
\[\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\dfrac{d}{{dx}}\left( {\sqrt {2 + cosx} - 1} \right)}}{{\dfrac{d}{{dx}}\left( {{{(\pi - x)}^2}} \right)}}\]
Now we know that:-
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
 And the derivative of a constant is zero.
Hence applying this formula we get:-
\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\dfrac{1}{{2\sqrt {2 + cosx} }}.\dfrac{d}{{dx}}\left( {2 + cosx} \right) - 0}}{{2\left( {\pi - x} \right).\dfrac{d}{{dx}}\left( {\pi - x} \right)}}\]
Now we know that:-
\[
  \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} \\
  \dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x \\
 \]
Also, the derivative of a constant is zero.
Hence applying these formulas we get:-
\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\dfrac{1}{{2\sqrt {2 + cosx} }}.\left( {0 - \sin x} \right)}}{{2\left( {\pi - x} \right).\left( { - 1} \right)}}\]
Solving it further we get:-
\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\dfrac{{\sin x}}{{2\sqrt {2 + cosx} }}}}{{2\left( {\pi - x} \right)}}\]
Simplifying it further we get:-
\[
  \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sin x}}{{2\sqrt {2 + cosx} .\left( {2\left( {\pi - x} \right)} \right)}} \\
   \Rightarrow \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sin x}}{{4\sqrt {2 + cosx} .\left( {\pi - x} \right)}} \\
 \]
Now putting in the limit we get:-
\[\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \dfrac{{\sin \pi }}{{4\sqrt {2 + cos\pi } .\left( {\pi - \pi } \right)}}\]
Now we know that:-
\[\sin \pi = 0\]
Also, \[cos\pi = - 1\]
Hence putting in the values we get:-
\[\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \dfrac{0}{{4\sqrt {2 - 1} .\left( {\pi - \pi } \right)}}\]
Evaluating its value we get:-
\[\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \dfrac{0}{0}\]
Now since we again got an indeterminate form
Therefore we need to apply L’ hospital rule again
Hence differentiating numerator and denominator separately we get:-
\[\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\dfrac{d}{{dx}}\left( {\sin x} \right)}}{{\dfrac{d}{{dx}}\left( {4\sqrt {2 + cosx} .\left( {\pi - x} \right)} \right)}}\]
Now we know that:-
\[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\]
Also, we will apply product rule in the denominator which states that:-
\[\dfrac{d}{{dx}}\left( {ab} \right) = b.\dfrac{d}{{dx}}\left( a \right) + a.\dfrac{d}{{dx}}\left( b \right)\]
Hence on applying these formulas we get:-
\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\cos x}}{{\left( {\pi - x} \right)\dfrac{d}{{dx}}\left( {4\sqrt {2 + cosx} } \right) + \left( {4\sqrt {2 + cosx} } \right)\dfrac{d}{{dx}}\left( {\pi - x} \right)}}\]
Now applying the following formula
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Also, the derivative of a constant term is zero
We get:-
\[\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\cos x}}{{\left( {\pi - x} \right).\dfrac{4}{{2\sqrt {2 + cosx} }}.\dfrac{d}{{dx}}\left( {2 + cosx} \right) + \left( {4\sqrt {2 + cosx} } \right)\left( {0 - 1} \right)}}\]
Now we know that:-
\[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]
Also, the derivative of a constant term is zero
Hence applying these formulas we get:-
\[\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\cos x}}{{\left( {\pi - x} \right).\dfrac{4}{{2\sqrt {2 + cosx} }}.\left( {0 - \sin x} \right) - \left( {4\sqrt {2 + cosx} } \right)}}\]
Simplifying it further we get:-
\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\cos x}}{{\dfrac{{ - 4\sin x\left( {\pi - x} \right)}}{{2\sqrt {2 + cosx} }} - \left( {4\sqrt {2 + cosx} } \right)}}\]
Now taking the LCM and solving it further we get:-
\[\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\cos x}}{{\dfrac{{ - 4\sin x\left( {\pi - x} \right) - 4\left( 2 \right){{\left( {\sqrt {2 + cosx} } \right)}^2}}}{{2\sqrt {2 + cosx} }}}}\]
Simplifying it further we get:-
\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\cos x\left( {2\sqrt {2 + cosx} } \right)}}{{ - 4\sin x\left( {\pi - x} \right) - 8\left( {2 + \cos x} \right)}}\]
Now putting in the limit we get:-
\[\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \dfrac{{\cos \pi \left( {2\sqrt {2 + cos\pi } } \right)}}{{ - 4\sin \pi \left( {\pi - \pi } \right) - 8\left( {2 + \cos \pi } \right)}}\]
Now we know that:-
\[\sin \pi = 0\]
Also, \[cos\pi = - 1\]
Hence putting the values we get:-
\[\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \dfrac{{\left( { - 1} \right)\left( {2\sqrt {2 - 1} } \right)}}{{ - 4\left( 0 \right)\left( 0 \right) - 8\left( {2 - 1} \right)}}\]
Solving it further we get:-
\[ \Rightarrow
  \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \dfrac{{\left( { - 1} \right)\left( {2\sqrt 1 } \right)}}{{0 - 8\left( 1 \right)}} \\
   \Rightarrow \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \dfrac{{ - 2}}{{ - 8}} \\
   \Rightarrow \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sqrt {2 + cosx} - 1}}{{{{(\pi - x)}^2}}} = \dfrac{1}{4} \\
 \]
Now since the given function is continuous at \[x = \pi \]
Therefore, we will equate the value so obtained with the value of function when \[x = \pi \]
On equating we get:-
\[k = \dfrac{1}{4}\]

Therefore, option D is the correct option.

Note: Students should take note that:
A function f is continuous when, for every value a in its domain the function is defined i.e.
\[f\left( a \right)\] is defined and \[\mathop {lim}\limits_{x \to a} f\left( x \right) = f(a)\].
Also, we have to apply the L’ hospital rule until the indeterminate form gets removed on evaluating the limit.