Let the limits are given as \[\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\sin x} \right)=l\] and \[\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\tan x} \right)=m\], then
(a) l exists but m does not
(b) m exists but l does not
(c) l and m both exist
(d) neither l nor m exists

Hint: Use basic rules of limit and inverse functions to evaluate the limit of given functions.

Complete step-by-step answer:
We have the functions \[{{\sec }^{-1}}\left( \dfrac{x}{\sin x} \right)\] and \[{{\sec }^{-1}}\left( \dfrac{x}{\tan x} \right)\]. We have to evaluate the limit for both functions around the point \[x=0\].
We will begin by evaluating the limits for the function \[\dfrac{x}{\sin x}\] and \[\dfrac{x}{\tan x}\].
We observe that \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\sin x}\] is of the form \[\dfrac{0}{0}\].
An indeterminate form is an expression involving two functions whose limit can’t be determined solely from the limits of the individual functions. We will use L’Hopital Rule to evaluate the limit of such forms which states that if \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{0}{0}\] then we have \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}=\dfrac{f'\left( a \right)}{g'\left( a \right)}\].
Substituting \[f\left( x \right)=x,g\left( x \right)=\sin x\] in the above formula and evaluating the limit around \[x=0\], we get \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d\left( x \right)}{dx}}{\dfrac{d\left( \sin x \right)}{dx}}.....\left( 1 \right)\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}\] is such that \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=1,n=1\] in the above formula, we get \[\dfrac{d\left( x \right)}{dx}=1.....\left( 2 \right)\].
We know that differentiation of any function of the form \[y=\sin x\] is such that \[\dfrac{d\left( \sin x \right)}{dx}=\cos x.....\left( 3 \right)\].
Substituting equation \[\left( 2 \right)\] and \[\left( 3 \right)\] in equation \[\left( 1 \right)\], we get \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d\left( x \right)}{dx}}{\dfrac{d\left( \sin x \right)}{dx}}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{\cos x}=1\].
Thus, we have \[\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\sin x} \right)=\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( 1 \right)=0\].
Hence, we have \[l=0\].
We will now evaluate the limit for \[\dfrac{x}{\tan x}\].
We observe that \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\tan x}\] is of the form \[\dfrac{0}{0}\].
Hence, we will use L’Hopital Rule to evaluate the limit which states that if \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{0}{0}\] then we have \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}=\dfrac{f'\left( a \right)}{g'\left( a \right)}\].
Substituting \[f\left( x \right)=x,g\left( x \right)=\tan x\] in the above formula and evaluating the limit around \[x=0\], we get \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d\left( x \right)}{dx}}{\dfrac{d\left( \tan x \right)}{dx}}.....\left( 4 \right)\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}\] is such that \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=1,n=1\] in the above formula, we get \[\dfrac{d\left( x \right)}{dx}=1.....\left( 5 \right)\].
We know that differentiation of any function of the form \[y=\tan x\] is such that \[\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x.....\left( 6 \right)\].
Substituting equation \[\left( 5 \right)\] and \[\left( 6 \right)\] in equation \[\left( 4 \right)\], we get \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\tan x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d\left( x \right)}{dx}}{\dfrac{d\left( \tan x \right)}{dx}}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{{{\sec }^{2}}x}=1\].
Thus, we have \[\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\tan x} \right)=\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( 1 \right)=0\].
Hence, we have \[m=0\].
We observe that both the limits exist and are equal to 0, i.e., \[l=m=0\], which is option (c).
Note: It’s necessary to evaluate the limit of the basic functions involved instead of directly evaluating the exact limit of the function. Indeterminate forms of the functions include \[\dfrac{0}{0},\dfrac{\infty }{\infty },0\times \infty ,\infty -\infty ,{{0}^{0}},{{1}^{\infty }},{{\infty }^{0}}\].