Limit Evaluation - 2

Consider the Taylor expansion:

$f(a+h) = f(a) + hf'(a) + \frac{h^2 f''(a)}{2!} + \dots$

$\therefore \tan^{-1}(a + h) \approx \tan^{-1}{a} + \frac{h}{1+a^2} - \frac{h^2a}{(1+a^2)^2}$

The numerator of the given limit is:

$N = 10\tan^{-1}(a -x) -15\tan^{-1}{a} -4\tan^{-1}(a +x) + 14\tan^{-1}(a +2x) - 6\tan^{-1}(a +3x) + \tan^{-1}(a +4x)$

Plugging the Taylor approximation of the arctangent function into the expression and simplifying leads to:

$N = -\frac{24x^2a}{(1+a^2)^2}$

The required limit therefore evaluates to:

$\lim_{x \to 0} \frac{N}{12x^2} = -\frac{2a}{(1+a^2)^2}$

One can plug in $a=2$ and take the absolute value of the fraction and get the required answer of $\boxed{29}$ . This expression is essentially the second derivative of the arctangent function $f(x) = \tan^{-1}{x}$ evaluated at $x=2$ . This is what the limit generally computes. You can use this limit to calculate second derivatives of other well known continuously differentiable functions too.

This method is clearly not the best one, as the second derivative is required to be computed to compute the second derivative using first principles, which is a weird thing to do logically. I look forward to seeing better approaches to this problem.