Limit of trig functions

Question

We have to evaluate $$\lim_{x\to 2} \frac{\cos^x a +\sin^x a -1}{x-2}.$$

I am working on it for hours

I tried using series , replacing $\cos a$ by $t$ and $\sin a$ by $\sqrt{1-t^2}$ but not got any result

How can I start it .

Answer 1

We need to assume that both $\cos a, \sin a$ are positive in order to evaluate this limit. This means that $a$ is in first quadrant. Assuming that this is so, we can simply assume that $\cos a = A, \sin a = B$ where $A, B$ are fixed positive numbers with $A^{2} + B^{2} = 1$ we can put $x = 2 + h$ so that as $x \to 2$ we have $h \to 0$. Then we have \begin{align} L &= \lim_{x \to 2}\frac{A^{x} + B^{x} - 1}{x - 2}\notag\\ &= \lim_{x \to 2}\frac{A^{2 + h} + B^{2 + h} - A^{2} - B^{2}}{h}\notag\\ &= \lim_{h \to 0}\frac{A^{2}\{\exp(h\log A) - 1\} + B^{2}\{\exp(h\log B) - 1\}}{h}\notag\\ &= \lim_{h \to 0}A^{2}\log A\frac{\{\exp(h\log A) - 1\}}{h\log A} + B^{2}\log B\frac{\{\exp(h\log B) - 1\}}{h\log B}\notag\\ &= A^{2}\log A + B^{2}\log B\notag\\ &= \cos^{2}a\log\cos a + \sin^{2}a\log\sin a\notag \end{align}