How can I solve $\lim_{x\to a}\; \frac{\cot(x)-\cot(a)}{x-a} $ without using L'Hopital's rule?

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Through some steps I got to this:

$$\lim_{x\to a}\; \frac{\frac{\cos(x)\sin(a)-\cos(a)\sin(x)}{x-a}}{\sin(x)\sin(a)}$$

The denominator is fine because if I place in $a$, I get $\sin^2(a)$. But the numerator should be $-1$ which I can't calculate without using the rule. I can't continue from here.

Thanks in advance

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$$\lim_{x\to a}\frac{\cot x-\cot a}{x-a}=(\cot x)'|_{x=a}=-\frac1{\sin^2 a}=-\csc^2(a)$$

Another way:

$$\frac{\cot x-\cot a}{x-a}=\frac{\frac{\cos x}{\sin x}-\frac{\cos a}{\sin a}}{x-a}=\frac{\sin a\cos x-\sin x\cos a}{(x-a)\sin x\sin a}=$$

$$\stackrel{\text{Trig. Identity}}=-\frac{\sin(a-x)}{a-x}\cdot\frac1{\sin x\sin a}\xrightarrow[x\to a]{}\stackrel{\text{known limit+continuity}}(-1)\cdot\frac1{\sin^2a}=-\csc^2(a)$$