If limit given as $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{-nx}}+{{e}^{nx}}-2\cos \dfrac{nx}{2}-k{{x}^{2}}}{\left( \sin x-\tan x \right)}$, exists and finite. Then possible values of ‘n’ and ‘k’ is:
(a) k = 3, n = 3
(b) k = 3, n = -2
(c) k = 5, n = 2
(d) k = -5, n = 2

Hint: Use expansion of all the functions given here ie. ${{e}^{nx}},\text{ }{{e}^{-nx}},\text{ }\cos \dfrac{nx}{2},\text{ }\sin x$ and tan x. Use the given condition to find n and k i.e. the limit is finite and does exist.
Complete step-by-step answer:
We have given that $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{-nx}}+{{e}^{nx}}-2\cos \dfrac{nx}{2}-k{{x}^{2}}}{\left( \sin x-\tan x \right)}$ exists and finite and need to determine n =? and k =?
We have $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{-nx}}+{{e}^{nx}}-2\cos \dfrac{nx}{2}-k{{x}^{2}}}{\sin x-\tan x}.....\left( i \right)$
Let us put the limit $x\to 0$ directly to numerator and denominator, we get,
\[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{-nx}}+{{e}^{nx}}-2\cos \dfrac{nx}{2}-k{{x}^{2}}}{\sin x-\tan x}\]
\[=\dfrac{{{e}^{0}}+{{e}^{-0}}-2\cos 0-k\left( 0 \right)}{\sin \left( {{0}^{o}} \right)-\tan \left( {{0}^{o}} \right)}\]
Now, we know that
\[\begin{align}
  & {{e}^{o}}=1 \\
 & \cos {{0}^{o}}=1 \\
 & \sin {{0}^{o}}=1 \\
 & \tan {{0}^{o}}=1 \\
\end{align}\]
Hence, we get above limit as
\[\dfrac{1+1-2}{0-0}=\dfrac{0}{0}\]
Hence, the expression is in indeterminate form after applying the limit \[x\to 0\].
Now, we need to use any other method to first remove the indeterminate form and then try to apply limits.
Now, as we can use the expansion of series \[{{e}^{x}},\cos x,\sin x\] and tan x very clearly to the equation (i) and try to get results in algebraic form.
Hence, we have the expansion of \[{{e}^{x}}\] as
\[{{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+.......\infty \]
Putting x = nx and x = – nx , we get
\[{{e}^{nx}}=1+nx+\dfrac{{{\left( nx \right)}^{2}}}{2!}+\dfrac{{{\left( nx \right)}^{3}}}{3!}+......\infty ....\left( ii \right)\]
\[{{e}^{-nx}}=1+\left( -nx \right)+\dfrac{{{\left( -nx \right)}^{2}}}{2!}+\dfrac{{{\left( -nx \right)}^{3}}}{3!}+......\infty \]
\[{{e}^{-nx}}=1-nx+\dfrac{{{\left( nx \right)}^{2}}}{2!}-\dfrac{{{\left( nx \right)}^{3}}}{3!}+......\infty ....\left( iii \right)\]
We have the expansion of cos x as
\[\cos x=1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}-\dfrac{{{x}^{6}}}{6!}+.....\infty \]
Putting \[x=\dfrac{nx}{2}\] which we require in question, we get
\[\cos \dfrac{nx}{2}=1-\dfrac{{{\left( \dfrac{nx}{2} \right)}^{2}}}{2!}+\dfrac{{{\left( \dfrac{nx}{2} \right)}^{4}}}{4!}-\dfrac{{{\left( \dfrac{nx}{2} \right)}^{6}}}{6!}+.....\infty \]
\[\cos \dfrac{nx}{2}=1-\dfrac{{{\left( nx \right)}^{2}}}{{{2}^{2}}2!}+\dfrac{{{\left( nx \right)}^{4}}}{{{2}^{4}}4!}-\dfrac{{{\left( nx \right)}^{6}}}{{{2}^{6}}6!}+.....\infty .....\left( iv \right)\]
And, we have the expansion of sin x and tan x as
\[\sin x=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}-\dfrac{{{x}^{7}}}{7!}+.....\infty ....\left( v \right)\]
\[\tan x=x+\dfrac{{{x}^{3}}}{3}+\dfrac{2{{x}^{5}}}{15}+.....\infty ....\left( vi \right)\]
Putting all the expansion as written in equations (ii), (iii), (iv), (v) and (vi) in equation (i), we get
Let limit of equation (i) is L, we get,
\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{nx}}+{{e}^{-nx}}-2\cos \dfrac{nx}{2}-k{{x}^{2}}}{\sin x-\tan x}\]
\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+nx+\dfrac{{{\left( nx \right)}^{2}}}{2!}+\dfrac{{{\left( nx \right)}^{3}}}{3!}+..... \right)+\left( 1-nx+\dfrac{{{\left( nx \right)}^{2}}}{2!}-\dfrac{{{\left( nx \right)}^{3}}}{3!}+.... \right)-2\left( 1-\dfrac{{{\left( nx \right)}^{2}}}{{{2}^{2}}2!}+\dfrac{{{\left( nx \right)}^{4}}}{{{2}^{4}}4!}+.... \right)-k{{x}^{2}}}{\left( x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+..... \right)-\left( x+\dfrac{{{x}^{3}}}{3}+\dfrac{2{{x}^{5}}}{15}+...... \right)}\]
Simplifying the above expression, we get
\[L=\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{2\left( 1+\dfrac{{{\left( nx \right)}^{2}}}{2!}+\dfrac{{{\left( nx \right)}^{4}}}{4!}+..... \right)-2\left( 1-\dfrac{{{\left( nx \right)}^{2}}}{{{2}^{2}}2!}+\dfrac{{{\left( nx \right)}^{4}}}{{{2}^{4}}4!}+.... \right)-k{{x}^{2}}}{{{x}^{3}}\left( \dfrac{-1}{3!}-\dfrac{1}{3} \right)+{{x}^{5}}\left( \dfrac{1}{5!}-\dfrac{2}{15} \right)+.....}\]
Cancelling out +2 and -2, we get
\[L=\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{2\left( \dfrac{{{\left( nx \right)}^{2}}}{2!}+\dfrac{{{\left( nx \right)}^{4}}}{4!}+..... \right)-2\left( \dfrac{-{{\left( nx \right)}^{2}}}{{{2}^{2}}2!}+\dfrac{{{\left( nx \right)}^{4}}}{{{2}^{4}}4!}+.... \right)-k{{x}^{2}}}{\dfrac{-{{x}^{3}}}{2}+{{x}^{5}}\left( \dfrac{1}{5!}-\dfrac{2}{15} \right)+.....}\]
Taking \[{{x}^{2}}\] common from numerator and denominator and after cancelling out them, we get
\[L=\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{2\left( \dfrac{{{n}^{2}}}{2!}+\dfrac{{{n}^{4}}{{x}^{2}}}{4!}+..... \right)+\dfrac{2{{n}^{2}}}{{{2}^{2}}2!}-\dfrac{2{{n}^{4}}{{x}^{2}}}{{{2}^{4}}4!}+....-k}{\dfrac{-x}{2}+{{x}^{3}}\left( \dfrac{1}{5!}-\dfrac{2}{15} \right)+.....}.....\left( vii \right)\]
Now, applying the limits to the numerator and denominator, we get
\[L=\text{ }\dfrac{\dfrac{2{{n}^{2}}}{2!}+0+\dfrac{{{n}^{2}}.2}{4.2}-0-k}{0}\]
\[L=\dfrac{{{n}^{2}}+\dfrac{{{n}^{2}}}{4}-k}{0}\]
\[L=\dfrac{\dfrac{5}{4}{{n}^{2}}-k}{0}\]
Now, we can observe that L will become positive infinite or negative infinite depending on the values of \[\dfrac{5{{n}^{2}}}{4}-k\]. If the value of \[\dfrac{5{{n}^{2}}}{4}-k\] is greater than 0, L will become positive infinite and if \[\dfrac{5{{n}^{2}}}{4}-k\] is less than 0, L will become negative infinite. But we need to observer that if \[\dfrac{5{{n}^{2}}}{4}-k\] will become 0, then L will become indeterminate form \[\left( \dfrac{0}{0} \right)\]. So, we will not be able to put \[x\to 0\] to the given expression if \[\dfrac{5{{n}^{2}}}{4}-k=0\], as the expression will be in indeterminate form. And we need to solve the expression or simplify it further for putting limits and to get any definite value. And it is given in the problem that L is a definite value and exists as well.
So, we need to put \[\dfrac{5{{n}^{2}}}{4}-k=0\], to get and indeterminate form of limit and we can solve ‘L’ further to get any definite value of it otherwise ‘L’ will become positive infinity or negative infinite if \[\dfrac{5{{n}^{2}}}{4}-k\] will take any value except ‘0’. Hence,
\[\dfrac{5{{n}^{2}}}{4}-k=0....\left( viii \right)\]
Hence, we can put \[\dfrac{5{{n}^{2}}}{4}-k=0\] in the equation (vii) and we get the value of ‘L’ as\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( \dfrac{{{n}^{4}}{{x}^{2}}}{4!}+.... \right)-\dfrac{{{n}^{4}}{{x}^{2}}}{{{2}^{4}}4!}+.....}{\dfrac{-x}{2}+{{x}^{3}}\left( \dfrac{1}{5!}-\dfrac{2}{15} \right)}\]
\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x\left[ \left( \dfrac{{{n}^{4}}x}{4!}+.... \right)-\dfrac{{{n}^{4}}x}{{{2}^{4}}4!}+..... \right]}{x\left[ \dfrac{-1}{2}+{{x}^{2}}\left( \dfrac{1}{5!}-\dfrac{2}{15} \right)+.... \right]}\]
Canceling ‘x’ from the denominator and numerator, we get,
\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( \dfrac{{{n}^{4}}x}{4!}+.... \right)-\dfrac{{{n}^{4}}x}{{{2}^{4}}4!}+.....}{\dfrac{-1}{2}+{{x}^{2}}\left( \dfrac{1}{5!}-\dfrac{2}{15} \right)+....}\]
Put \[x\to 0\] to the above expression, we get the value of ‘L’ as
\[L=\dfrac{0-0}{\dfrac{-1}{2}+0}=0\]
Hence, putting \[x\to 0\] to the expression, we get L = 0. Hence, the limit \[x\to 0\] to the given expression is 0.
Now, we have only one relation between n and k i.e.
\[\dfrac{5}{4}{{n}^{2}}-k=0.....\left( ix \right)\]
Hence, there are infinite values possible for n and k.
For option (a) k = 3, n = 3
Putting LHS of equation (ix), we get
\[\dfrac{5}{4}\times {{3}^{2}}-3=\dfrac{45}{4}-3\ne 0\ne RHS\]
Hence, it’s not true.
For option (b) k = 3, n = -2
Putting to LHS of equation (ix), we get
\[\dfrac{5}{4}\times {{\left( -2 \right)}^{2}}-3=5-3\ne 0\ne RHS\]
Hence, it’s not true.
For option (c) k = 5, n = 2,
Putting in equation (ix), we get
\[\dfrac{5}{4}\times {{\left( 2 \right)}^{2}}-5=5-5=0=RHS\]
Hence, it is the correct answer.
For option (d) k = -5, n = 2
Putting in equation (ix), we get
\[\dfrac{5}{4}\times {{\left( 2 \right)}^{2}}+5=10\ne 0\ne RHS\]
Hence, it’s not true.
Hence, option (c) is the correct answer from the given options.

Note: Another approach for this question would be that we can use L’Hospital theorem to simplify the given relation as:
$L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{-nx}}+{{e}^{nx}}-2\cos \dfrac{nx}{2}-k{{x}^{2}}}{\left( \sin x-\tan x \right)}$
Differentiating numerator and denominator as
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{-n{{e}^{-nx}}+n{{e}^{nx}}+n\sin \dfrac{nx}{2}-2kx}{\cos x-{{\sec }^{2}}x}$
Now, we can put the limit $x\to 0$ to the expression. But we will get $\dfrac{0}{0}$ form now as well. Hence, again differentiating the expression, we get
\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{n}^{2}}{{e}^{-nx}}+{{n}^{2}}{{e}^{nx}}+\dfrac{{{n}^{2}}}{2}\cos \dfrac{nx}{2}-2k}{-\sin x-2{{\sec }^{2}}x\tan x}\]
Now, putting the limit $x\to 0$ to the given expression we get
\[L=\dfrac{{{n}^{2}}+{{n}^{2}}+\dfrac{{{n}^{2}}}{2}-2k}{0}\]
\[L=\dfrac{\dfrac{5{{n}^{2}}}{2}-2k}{0}\]
Now, we know that limit is definite, hence \[\dfrac{5{{n}^{2}}}{2}-2k\] should be zero to form indeterminate form, so that one can simplify the relation, then put limits to it.
Hence, \[\dfrac{5{{n}^{2}}}{2}=2k\]
Or, \[\dfrac{5{{n}^{2}}}{4}-k=0\]
Same relation as in the solution.
Now, after differentiation again, we will get limit L to zero.
One can get confused in the coefficients of ‘x’ that is why we did not take high power of ‘x’ in the expansions. The reason is simple, they will become zero once the limit is put in the expression.