<s>[INST] How to evaluate $\lim\limits_{x\to y}\frac{\tan x-\tan y}{x-y}$? 
Evaluate $$\lim_{x\to y}\frac{\tan x-\tan y}{x-y}.$$

I used elementary methods to evaluate the limit. Here is my solution:
Let $d=x-y$. Then as $x\to y$, we have $d\to 0$. Now,
$\lim\limits_{x\to y}\frac{\tan x-\tan y}{x-y}\\ =\lim\limits_{d\to 0}\frac{\tan x-\tan (x-d)}d\\ =\lim\limits_{d\to 0}\frac{\tan x-\frac{\tan x-\tan d}{1+\tan x\tan d}}{d}\\ =\lim\limits_{d\to 0}\frac{\tan x(1+\tan x\tan d)-\tan x+\tan d}{d(1+\tan x\tan d)}\\ =\lim\limits_{d\to 0}\frac{\tan x(1+\tan x\tan d-1)}{d(1+\tan x\tan d)}+\frac{\tan d}d \cdot \frac1{1+\tan x\tan d}\\ =\lim\limits_{d\to 0}\frac{\tan^2x}{1+\tan x\tan d}\cdot \frac{\tan d}d+\frac{\tan d}d \cdot \frac1{1+\tan x\tan d}\\ =\lim\limits_{d\to 0}\frac{\tan^2x}{1+\tan x\tan 0}\cdot 1+1\cdot\frac1{1+\tan x\tan 0}\\ =\tan^2x+1$
I want to know whether my solution is correct or not. And can this limit be evaluated using L'Hôpital's rule? (I've just learnt L'Hôpital's rule and I don't know how to use this to evaluate a limit which contains two variables). And some other methods to solve the problem are also welcome.
 [/INST] You can use the definition of the derivative to evaluate such a limit :
$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)$$
But here's an approach if you want to exercise more, by the end of this answer I'll mention the important properties you have to know :
\begin{align} 
\lim_{x\to y} \frac{\tan x -\tan y}{x-y}&=\lim_{x\to y} \frac{\frac{\sin x}{\cos x}-\frac{\sin y}{\cos y}}{x-y}\\
&=\lim_{x\to y} \frac{\sin x \cos y-\sin y\cos x}{\cos y\cos a}\frac{1}{x-y} \\
&=\lim_{x\to y} \frac{\frac{1}{2}\left(\sin(x+y)+\sin(x-y) -\sin(y+x)-\sin(x-y)\right)}{\frac{1}{2}\left(\cos(x+y)+\cos(x-y)\right)}\frac{1}{x-y} \\
&=\lim_{x\to y} \frac{-2\sin(y-x)}{\cos(2y)+1}\frac{-1}{y-x}\\
&=\lim_{x\to y} \frac{\sin (y-x)}{y-x} \frac{1}{\cos^2 y}\\
&= \frac{1}{\cos^2 y}
\end{align}

Recall :
$$\sin \alpha \cos \beta =\frac{1}{2} \left(\sin (\alpha -\beta)+\sin(\alpha +\beta)\right)$$
$$\cos \alpha \cos \beta =\frac{1}{2} \left(\cos(\alpha -\beta)+\cos(\alpha+\beta)\right)$$
And this usual limit :
$$\lim_{\alpha\to 0}\frac{\sin(\alpha)}{\alpha}=1$$
Note that :
$$\frac{1}{\cos^2 \alpha}=\frac{\cos^2 \alpha +\sin^2 \alpha}{\cos^2 \alpha} 
=1+\frac{\sin^2 \alpha}{\cos^2\alpha}
=1+\tan^2 \alpha $$
Good luck !
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