Let \(f\left( x \right) = \frac{{\sin \left( x \right)}}{{x - 54}}\). Then f⁽¹⁰⁰⁾(54) is given by

Concept:

Taylor’s series method:

The Taylor series can be used to calculate the value of an entire function at every point, if the value of the function, and all of its derivatives, are known at a single point.

Taylor's series expansion for f (x + h) is

\(f\left( {x + h} \right) = f\left( x \right) + hf'\left( x \right) + \frac{{{h^2}}}{{2!}}f''\left( x \right) + \frac{{{h^3}}}{{3!}} + f'''\left( x \right) + \ldots \infty \)

\(f(x)=f(a)\;+\;(x-a)f'(x)\;+\;\frac{(x-a)^2}{2!}f''(x)\;+\;........\infty\)

Calculation:

Given:

\(f\left( x \right) = \frac{{\sin \left( x \right)}}{{x - 54}}\)

f(100)(54) = ?

Using Taylor series expansion for Sin x at a = 54

\(f(x)=f(a)\;+\;(x-a)f'(x)\;+\;\frac{(x-a)^2}{2!}f''(x)\;+\;........\infty\)

\(\sin x~=~\sin (54)~+~\frac{(x~-~54)~\times~\cos (54)}{1!}~-~\frac{(x~-~54)^2~\times~\cos (54)}{2!}~-~.............\infty\)

Now the function transforms into:

f(x) = \(\frac{\sin (54)}{x~-~54}~+~\frac{\cos (54)}{1!}~-~\frac{(x~-~54)~\times~\cos (54)}{2!}~-~.............\infty\)

After Observing carefully the first term in the above infinite series, the (x - 54) term is always in the denominator, which will become zero when we put x = 54.

Every derivative will also have the same term till infinite.

So, every term will have zero in its denominator after putting x = 54.

⇒ f(100)(54) is Undefined.