The graph of the function $y=f\left( x \right)$ has a unique tangent at $\left( {{e}^{a}},0 \right)$ through which the graph passes, then $\displaystyle \lim_{x \to {{e}^{a}}}\dfrac{\log \left( 1+7f\left( x \right) \right)-\sin \left( f\left( x \right) \right)}{3f\left( x \right)}$
A. 1
B. 2
C. 7

Hint: We are required to know the basic derivatives formulae in order to solve this question. We first substitute the limits directly and try to calculate the value for the output. We get a $\dfrac{0}{0}$ form. We use L'Hospital's rule in order to eliminate this $\dfrac{0}{0}$ form. This is done by differentiating the numerator and denominator and writing the fraction. We now substitute the value of limit and we get our answer.

Complete step by step solution:
Given the function $y=f\left( x \right),$ we know that since there exists a tangent at $\left( {{e}^{a}},0 \right)$ ,we can say that this point satisfies the function. This is shown by substituting the point $\left( {{e}^{a}},0 \right)$ in the function $y=f\left( x \right),$
$= 0=f\left( {{e}^{a}} \right)$
Also, by differentiating the function and substituting the same point,
${{\left. \dfrac{dy}{dx} \right|}_{\left( {{e}^{a}},0 \right)}}=f'\left( {{e}^{a}} \right)$
Now considering the question,
$= \displaystyle \lim_{x \to {{e}^{a}}}\dfrac{\log \left( 1+7f\left( x \right) \right)-\sin \left( f\left( x \right) \right)}{3f\left( x \right)}$
Substituting the limits directly in place of x,
$= \dfrac{\log \left( 1+7f\left( {{e}^{a}} \right) \right)-\sin \left( f\left( {{e}^{a}} \right) \right)}{3f\left( {{e}^{a}} \right)}$
We know that $f\left( {{e}^{a}} \right)=0,$
$= \dfrac{\log \left( 1+7.0 \right)-\sin \left( 0 \right)}{3.0}$
We know that $\log 1,$ $\sin 0,$ and the denominator term 3 multiplied by 0 all becomes zero.
Therefore, we have
$= \dfrac{\log \left( 1+7.0 \right)-\sin \left( 0 \right)}{3.0}=\dfrac{0}{0}$
This gives us an indeterminate form and we need to use L Hospital’s Rule to obtain a finite solution. This is done by differentiating the numerator and denominator separately and writing them in the same form. We need to differentiate the $\log \left( 1+7f\left( x \right) \right)$ by using the differentiation of a composite method. We know the derivative of $\sin \left( f\left( x \right) \right)$ can again be done by the differentiation of a composite method. Differentiation of $f\left( x \right)$ in denominator gives us $f'\left( x \right).$ Performing all this,
$= \displaystyle \lim_{x \to {{e}^{a}}}\dfrac{\dfrac{1}{1+7f\left( x \right)}.7f'\left( x \right)-\cos \left( f\left( x \right) \right).f'\left( x \right)}{3f'\left( x \right)}$
Taking all the $f'\left( x \right)$ common out from numerator and denominator,
$= \displaystyle \lim_{x \to {{e}^{a}}}\dfrac{f'\left( x \right).\left( \dfrac{7}{1+7f\left( x \right)}-\cos \left( f\left( x \right) \right) \right)}{3.f'\left( x \right)}$
Cancelling the $f'\left( x \right)$ term and substituting the limits,
$= \dfrac{\dfrac{7}{1+7f\left( {{e}^{a}} \right)}-\cos \left( f\left( {{e}^{a}} \right) \right)}{3}$
We know that $f\left( {{e}^{a}} \right)=0,$
$= \dfrac{\dfrac{7}{1+7.0}-\cos \left( 0 \right)}{3}$
We know that $\cos 0=1,$ and we substitute this in the above equation,
$= \dfrac{7-1}{3}$
Subtracting 1 from 7 and dividing by 3 we get,
$= \dfrac{7-1}{3}=2$

So, the correct answer is “Option B”.

Note: It is important to know the basic derivation formulae and the usage of L Hospital’s rule in order to simplify limits producing an indeterminate form. Here, we use the method of differentiation of composite function where, we first differentiate the outside function and multiply by the differentiation of the inner function.