Applying Lagrange’s Mean Value Theorem for a suitable function f (x) in [0, h], we have \[f\left( h \right)=f\left( 0 \right)+hf'\left( \theta h \right)\], \[0<\theta <1\]. Then for \[f\left( x \right)=\cos x\], the value of \[\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\theta \] is
(a) 1
(b) 0
(c) \[\dfrac{1}{2}\]
(d) \[\dfrac{1}{3}\]

Hint:By Lagrange’s theorem, interval [0, h] is differentiable on an open interval (0, h). From the given expression find the value of f (h), f (0) and \[f'\left( \theta h \right)\] by using the function, \[f\left( x \right)=\cos x\]. Substitute back the values, simplify it and apply \[\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\] to the expression and get the value of limit.

Complete step-by-step answer:
Lagrange’s mean value theorem states that if a function f (x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x = c on this interval such that,
\[f'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}\]
This theorem allows to express the increment of a function and can interval through the value of the derivative at an intermediate point of the segment.
Now from Lagrange’s mean value theorem, there exist \[c\in \left( a,b \right)\] such that,
\[f'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}.......(1)\]
We have been given the closed interval, [a, b] = [0, h].
Thus, we can say that, a = 0 and b = h.
We have also been given that,
\[f\left( h \right)=f\left( 0 \right)+hf'\left( \theta h \right)\]
Let us rearrange it as, \[hf'\left( \theta h \right)=f\left( h \right)-f\left( 0 \right).....(2)\]
Now, put \[c=\theta h\], b = h and a = 0 in (1), we get,
\[f'\left( \theta h \right)=\dfrac{f\left( h \right)-f\left( 0 \right)}{h-0}\]
\[\therefore f'\left( \theta h \right)=\dfrac{f\left( h \right)-f\left( 0 \right)}{h}......(3)\]
We have been given that, \[f\left( x \right)=\cos x\]
\[\begin{align}
  & \therefore f\left( h \right)=\cosh \\
 & f\left( 0 \right)=\cos 0=1 \\
\end{align}\]
\[f\left( x \right)=\cos x\], differentiating it w.r.t x
\[f'\left( x \right)=-\sin x\]
Similarly, \[f'\left( \theta h \right)=-\sin \left( \theta h \right)\]
Now let us put the values of \[f'\left( \theta h \right)\], f (h) and f (0) in (3).
\[-\sin \left( \theta h \right)=\dfrac{\cosh -1}{h}\]
The cosine function has the power series expansion as,
\[\begin{align}
  & \cos x=\sum\limits_{x=0}^{\infty }{{{\left( -1 \right)}^{n}}}\dfrac{{{x}^{2n}}}{\left( 2n \right)!} \\
 & \cos x=1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}-\dfrac{{{x}^{6}}}{6!}+.... \\
\end{align}\]
Similarly, \[\cosh =1-\dfrac{{{h}^{2}}}{2!}+\dfrac{{{h}^{4}}}{4!}-\dfrac{{{h}^{6}}}{6!}+....\]
Hence let us put \[\cosh =\left( 1-\dfrac{{{h}^{2}}}{2!} \right)\], neglecting other higher powers of h.
\[\therefore -\sin \left( \theta h \right)=\dfrac{\left( 1-\dfrac{{{h}^{2}}}{2!} \right)-1}{h}\], let us simplify it
\[-\sin \left( \theta h \right)=\dfrac{1-\dfrac{{{h}^{2}}}{2!}-1}{h}\Rightarrow -\sin \left( \theta h \right)=\dfrac{-{{h}^{2}}}{2!}\times \dfrac{1}{h}\]
\[\therefore \sin \left( \theta h \right)=\dfrac{h}{2}\]
i.e. \[\theta h={{\sin }^{-1}}\left( \dfrac{h}{2} \right)\]
\[\therefore \theta =\dfrac{{{\sin }^{-1}}\dfrac{h}{2}}{h}\], now multiply by \[\left( \dfrac{1}{2} \right)\] in the numerator and the denominator. We get,
\[\begin{align}
  & \theta =\dfrac{{{\sin }^{-1}}\dfrac{h}{2}\times \dfrac{1}{2}}{h\times \dfrac{1}{2}} \\
 & \therefore \theta =\dfrac{{{\sin }^{-1}}\dfrac{h}{2}\times \dfrac{1}{2}}{\dfrac{h}{2}} \\
\end{align}\]
We know that, \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1\].
Similarly, \[\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\theta =\dfrac{1}{2}\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}\dfrac{h}{2}}{\dfrac{h}{2}}=\dfrac{1}{2}\times 1=\dfrac{1}{2}\]
Thus we got, \[\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\theta =\dfrac{1}{2}\].
\[\therefore \] Option (c) is the correct answer.

Note: From the given function, \[f\left( h \right)=f\left( 0 \right)+hf'\left( \theta h \right)\], by rearranging it we will get the Lagrange’s Mean value formula. In order to understand it better, you should know the theorem and then apply basic identities and solve it.Students should also remember the power series expansion of $\cos x$ i.e $\cos x=1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}-\dfrac{{{x}^{6}}}{6!}+....$ to solve these types of questions.