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.
Best Answer
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 $$ f'(y)=\dfrac{\cos^2y+\sin^2y}{\cos^2y}=1+\tan^2y $$