I have to calculate this limit whitout using L'Hopital's rule or Taylor polynomials:
$$\lim_{ x\to \pi/4 } \frac{1 - \tan(x)}{x-\frac{\pi}{4}}$$
I know how to make it using L'Hopital and that the result is $-2$ ,but I'm getting nowhere when I try without it. Any advice?
$\endgroup$ 05 Answers
$\begingroup$Hint: What is the definition of the derivative of $\tan(x)$ at $x=\pi/4$?
$\endgroup$ 6 $\begingroup$Recall that $$ f'(a) = \lim_{x\to a}\frac{f(x) - f(a)}{x-a}. $$ Apply it to the case where $f(x) = \tan x$ and $a=\pi/4$. Then $f(a) = 1$ and $f'(a) = \sec^2 a = \sec^2(\pi/4) = 2$. Therefore $$ 2 = \lim_{x\to\pi/4}\frac{\tan x - \tan(\pi/4)}{x-\pi/4}. $$
So $-2$ is the answer to the question as you've posed it.
$\endgroup$ $\begingroup$We have the identities
$$\begin{align}\frac{1-\tan x}{x-\pi/4}&=-\frac{\tan(x-\pi/4)}{x-\pi/4}(1+\tan x)\\\\ &=-\frac{\sin(x-\pi/4)}{x-\pi/4}\frac{1+\tan x}{\cos(x-\pi/4)} \end{align}$$
Can you finish from here?
$\endgroup$ 2 $\begingroup$Hint:
Let $t=x-\frac{\pi}{4}$, then $$\frac{1-\tan x}{x-\frac{\pi}{4}}=\frac{1-\tan\left(t+\frac{\pi}{4}\right)}{t}=\frac{1}{t}\cdot\left(1-\frac{\tan t+\tan\frac{\pi}{4}}{1-\tan \frac{\pi}{4}\tan t}\right)=\frac{1}{t}\left(1-\frac{\tan t+1}{1-\tan t}\right)=\frac{-2\tan t}{t(1-\tan t)}$$ Now, take the limit as $t\to 0$: \begin{align} \lim_{x\to \frac{\pi}{4}}\frac{1-\tan x}{x-\frac{\pi}{4}}&=\lim_{t\to 0}\frac{-2\tan t}{t(1-\tan t)}\\ &=\lim_{t\to 0}\frac{-2\tan t\cos t}{t(1-\tan t)\cos t}\\ &=-2\lim_{t\to 0}\frac{\sin t}{t(\cos t-\sin t)}\\ &=-2\left(\lim_{t\to 0}\frac{\sin t}{t}\right)\left(\lim_{t\to 0}\frac{1}{\cos t-\sin t}\right)\\ &=-2(1)(1)\\ &=\color{blue}{-2} \end{align}
$\endgroup$ 1 $\begingroup$You could use the following hint. Try to write $\tan(x)$ in terms of $\sin(x) $ and $\cos(x)$. A little bit of rearranging and then use the following,
$$\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1$$
A little bit of work yields the result.
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