# Calculate limit without L'Hopital's rule

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?

Hint: What is the definition of the derivative of $\tan(x)$ at $x=\pi/4$?

• $\frac{1}{\cos^2(\frac{\pi}{4})}$ But what's the point? I can't use L'Hopital's rule because the exercise ask it, not because I don't how to use it.I don't understand how can I use the derivative of $\tan(x)$ at $\frac{\pi}{4}$ in this exercise. Commented Sep 30, 2015 at 21:47
• @DerRosenkavalier By definition: $$\tan'(\pi/4)=\displaystyle\lim_{x \to \pi/4}\frac{\tan(x)-1}{x-\pi/4}$$. How does this compare to your limit? Commented Sep 30, 2015 at 21:49
• L'Hopital isn't being used here; just the definition of the derivative. It's important to know the difference.
– zhw.
Commented Sep 30, 2015 at 22:00
• @zhw I agree with your comment. This type of question sometimes rasises debate as to "allowed ways forward. For example, is the use of asymptotic analysis permitted here? IMHO, yes absolutely. Yet others would argue that that approach is tantamount to the use of LHR. ;-) Commented Sep 30, 2015 at 22:21
• @DerRosenkavalier : My answer posted here explicitly explains how you can use the value of the derivative of $\tan x$ at $\pi/4$. ${}\qquad{}$ Commented Oct 1, 2015 at 4:32

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.

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}

• While all the answers are valid and helped me,I chose this because it doesn't imply the use of more advanced concepts like derivatives.Thanks! Commented Oct 4, 2015 at 20:48

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?

• Yes,I could use that $\lim_{ x\to 0 }\frac{\sin(x)}{x}=1$ . Thanks for the help! Commented Oct 4, 2015 at 20:58
• You're welcome. My pleasure. This is a very efficient approach that circumvents use of L'Hospital's Rule, as you requested. Commented Oct 4, 2015 at 21:07

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.