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.
 A: The derivation is incorrect, because
$$
\lim_{x\to y}\frac{\tan x-\tan y}{x-y}
$$
cannot depend on $x$, which is a dummy variable. The limit, if it exists, is the same as
$$
\lim_{t\to y}\frac{\tan t-\tan y}{t-y}
$$
Where did you go wrong? When you do $d=x-y$, you need to substitute $x=y+d$, to get
$$
\lim_{d\to 0}\frac{\tan(y+d)-\tan y}{d}
$$
Then you can go on like you did:
\begin{align}
\lim_{d\to 0}\frac{\tan(y+d)-\tan y}{d}
&=\lim_{d\to0}\frac{\dfrac{\tan y+\tan d}{1-\tan y\tan d}-\tan y}{d} \\[6px]
&=\lim_{d\to 0}\frac{\tan y+\tan d-\tan y+\tan^2y\tan d}{d} \\[6px]
&=\lim_{d\to0}\frac{\tan d}{d}(1+\tan^2y) \\[6px]
&=\lim_{d\to0}\frac{\sin d}{d}\frac{1}{\cos d}(1+\tan^2y) \\[6px]
&=1+\tan^2y
\end{align}
Can you apply l’Hôpital? Not really. What you're computing is the derivative of the function $f(t)=\tan t$ at $y$ and in order to apply l’Hôpital you need to know this derivative: it’s circular, isn't it?
Of course you know the quotient rule, so, using $f(t)=\frac{\sin t}{\cos t}$, you have
$$
f'(y)=\dfrac{\cos^2y+\sin^2y}{\cos^2y}=1+\tan^2y
$$
A: 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 !
A: Usually when $a\to b$ it means that we consider a function of a variable $a$ at a fixed point $b$ - just in terms of easier interpretation. Hence, in your case $y$ is fixed and $x$ is a variable. As a result, you need to compute
$$
\lim\limits_{x\to y}\frac{f(x)}{g(x)}
$$
where $f(x) = \tan x - \tan y$ and $g(x) = x - y$, here $y$ you treat as a parameter. Conditions of the theorem are met, hence you get
$$
\lim\limits_{x\to y}\frac{f(x)}{g(x)} = \lim\limits_{x\to y}\frac{f'(x)}{g'(x)} = f'(y)
$$
since $g'(x) = 1$ for all $x \in \Bbb R$.
A: One might argue that since $\tan x = \frac{\sin x}{\cos x}$, we can compute $(\tan x)'$ from the product rule and then substitute in the original limit
$$ \lim _{h\to 0} \frac{\tan (x+h)-\tan x}{h} \overset{L'H} =\lim _{h\to 0} \frac{1}{\cos ^2(x+h)} = \frac{1}{\cos^2x} \ldots\text{right?} $$
No! This is circular! Using the validity of the very thing we are trying to prove. This is how many students often try to weasel their way out of proving a limit.
You could use L'Hopital's rule, but you have to be careful of not actually using the derivative of $\tan x$ in it, since that's precisely what you are computing by definition in the first place.
