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I'm having a problem to solve this limit.

$$\lim_{x \to \pi/4} \frac{\tan x-1}{\sin x-\cos x}$$

$\lim_{x \to \pi/4} \frac{\tan x-1}{\sin x-\cos x}$ = $\lim_{x \to \pi/4} \frac{\frac{\sin x}{\cos x}-1}{\sin x-\cos x}$= $\lim_{x \to \pi/4} \frac{\frac{\sin x-\cos x}{\cos x}}{\sin x-\cos x}$= $\lim_{x \to \pi/4} \frac{\frac{\frac{\sin x-\cos x}{\cos x}}{\sin x-\cos x}}{1}$ =

numerator is : (upper*lower) = 1*$\sin x-\cos x$

denominator is : (inner-up*inner-low) = $\cos x*(\sin x-\cos x)$.

Which is :

$$\lim_{x \to \pi/4} \frac{\sin x-\cos x}{(\cos x)(\sin x-\cos x)}$$

I don't know what to do next? Any ideas?

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Please edit your question. Put a slash before the trigonometry function. For instance, \sin x. –  Tunk-Fey Jul 26 '14 at 18:21
Why is denominator on top? –  Karolis Juodelė Jul 26 '14 at 18:24
Edited, sorry confused. –  JaVaPG Jul 26 '14 at 18:31
Cancel $\sin x-\cos x$ from the numerator and denominator, getting $\dfrac{1}{\cos x}$, and go on from there. I don't why anybody would answer this by saying anything but that. –  Michael Hardy Jul 26 '14 at 22:31

2 Answers 2

up vote 3 down vote accepted

Rearrange the fraction into something simpler. Express $\tan(x)$ in terms of $\sin(x)$ and $\cos(x)$. $$\frac{\tan(x)-1}{\sin(x)-\cos(x)}=\frac{\frac{\sin(x)}{\cos(x)}-1}{\sin(x)-\cos(x)}$$ You've done this and the next step in your own work. $$=\frac{\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\cos(x)}}{\sin(x)-\cos(x)}$$ $$=\frac{\frac{\sin(x)-\cos(x)}{\cos(x)}}{\sin(x)-\cos(x)}$$ Factor out $(\sin(x)-\cos(x))$ from the numerator and denominator. $$=\frac{\sin(x)-\cos(x)}{\sin(x)-\cos(x)}\cdot\frac{\frac{1}{\cos(x)}}{1}$$ $$=\frac{\frac{1}{\cos(x)}}{1}=\frac{1}{\cos(x)}$$ $$\therefore\:\frac{\tan(x)-1}{\sin(x)-\cos(x)}=\frac{1}{\cos(x)}$$ Apply the limit to both sides. $$\lim_{x \to \pi/4} \frac{\tan x-1}{\sin x-\cos x}=\lim_{x \to \pi/4} \frac{1}{\cos(x)}$$ The limit is then the value of $\frac{1}{\cos(x)}$ at $x=\frac{\pi}{4}$. In this particular case, you can substitute $x=\frac{\pi}{4}$ (but you can't always do this). $$\frac{1}{\cos(\frac{\pi}{4})}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}$$ $$\therefore\:\sqrt{2}$$

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I don't understand the calculation, what is sec(x)? –  JaVaPG Jul 26 '14 at 18:37
$\sec(x)$ is $\frac{1}{\cos(x)}$; I've edited my answer to be clearer –  Eul Can Jul 26 '14 at 18:40
Can you explain how did you managed to get to $\frac{1}{\cos x}$ from the original fraction? –  JaVaPG Jul 26 '14 at 18:43
@JaVaPG Is it clearer now? It pretty much starts where you left off but I could add more steps if it looks confusing. –  Eul Can Jul 26 '14 at 18:55
I'll be very glad if you could add some more steps. –  JaVaPG Jul 26 '14 at 18:56

Safely cancel out $\sin x-\cos x$ as $\displaystyle x\to\frac\pi4,\sin x-\cos x\to0\implies\sin x-\cos x\ne0$

Things will be clearer if we write $$\lim_{x \to \pi/4} \frac{\tan x-1}{\sin x-\cos x}=\lim_{x\to\dfrac\pi4}\frac{\tan x-1}{\cos x(\tan x-1)}$$

As $\displaystyle x\to\frac\pi4,\tan x\to1\implies\tan x\ne1$

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$$\sin \pi/4 = \frac{\sqrt{2}}{2}$$ and also $$\cos \pi/4 = \frac{\sqrt{2}}{2}$$ Meaning that sinx-cosx = 0 –  JaVaPG Jul 26 '14 at 18:27
@JaVaPG,Please find the edited version –  lab bhattacharjee Jul 26 '14 at 18:32
Can you explain how $\sin x - \cos x$ = $\cos x(\tan x - 1)$? –  JaVaPG Jul 26 '14 at 18:38
@JaVaPG, $$\tan x-1=\frac{\sin x}{\cos x}-1=?$$ –  lab bhattacharjee Jul 26 '14 at 18:40
$$\tan x -1 = \frac{\sin x}{\cos x} - 1 = \frac{\sin x-\cos x}{\cos x}$$ However I don't quite understand since $$\tan \pi/4 = 1$$ so the denominator and numerator are equal to 0. –  JaVaPG Jul 26 '14 at 18:46

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