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I am trying to evaluate $$\lim_{x \to \frac{\pi}{4}} \frac{1-\tan x}{x-\frac{\pi}{4}}$$ without using L'hopital's rule. However, I am not sure what to do. The only thing that came to my mind was to change the tan to sin over cos and get a common denominator but I felt that won't get me anywhere. A hint will be greatly appreciated.

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    Set $1=\tan\pi/4$ or set $x-\pi/4=u$ – 2017-02-28
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    Your expression is $-\frac{\tan x-\tan(\pi/4)}{x-\pi/4}$. The limit is the derivative definition $-\tan'(\pi/4)$. – 2017-02-28
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    @labbhattacharje Ok. I got my expression to $\lim_{u \to 0} \frac{1-\tan (u+\frac{\pi}{4})}{u}$. I still don't see how I got closer – 2017-02-28
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    @AspiringMathlete Checkout my answer. – 2017-02-28
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    @AspiringMathlete, see http://math.stackexchange.com/questions/2165038/find-the-limit-without-lhopital-rule-lim-x-to-1-frac1-cot-fracπ4x/2165071#2165071 – 2017-02-28

4 Answers 4

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Hint

This kind of problems are rather simple to address if you know Taylor series. Assuming you do, built at $t=a$, you have

$$\tan(x)=\tan (a)+ \left(\tan ^2(a)+1\right)(x-a)+O\left((x-a)^2\right)$$

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    Sorry, I don't know Taylor series. – 2017-02-28
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    @AspiringMathlete. Don't worry ! You will learn about them very soon and I hope than you will share my passion for them (for more than 60 years now). Good luck in your studies. Cheers :-) – 2017-02-28
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We're looking for

$$\lim_{x\to{\pi\over 4}}-{\tan{x}-\tan{\pi\over 4}\over x-{\pi\over 4}}$$

And this is $-\tan'{\pi\over 4}=-2$

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    I don't understand how you got the tanx in the numerator. – 2017-02-28
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    $1=\tan{\pi\over 4}$ – 2017-02-28
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    Ok. I understand but since when are we allowed to substitute the value of limits into a limit we haven't evaluated yet? – 2017-02-28
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    @AspiringMathlete Where do you see that this suggests "to substitute the value of limits into a limit we haven't evaluated yet"? Please be specific. – 2017-02-28
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    @AspiringMathlete: S/he didn't, he just noted that the $1$ which was already there equals $\tan\frac\pi4$. – 2017-02-28
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    $\tan{\pi\over 4}=1$ does not depend on $x$ so there's no limit in there. – 2017-02-28
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If you substitute $t=x-\pi/4$, then $$ \tan x=\tan(t+\pi/4)=\frac{\tan t+1}{1-\tan t} $$ so your limit is $$ \lim_{t\to0}\frac{1}{t}\left(1-\frac{\tan t+1}{1-\tan t}\right)= \lim_{t\to0}\frac{\tan t}{t}\frac{-2}{1-\tan t} $$ Alternatively, recall that $$ \cos x-\sin x=-\sqrt{2}\sin\left(x-\frac{\pi}{4}\right) $$ and therefore $$ 1-\tan x=-\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)\frac{1}{\cos x} $$ Hence the limit can be rewritten as $$ \lim_{x\to\pi/4}\frac{\sin\left(x-\frac{\pi}{4}\right)}{x-\frac{\pi}{4}}\frac{-\sqrt{2}}{\cos x} $$

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    I see you used the identity of $\tan(a+b)$. I like this solution. Just out of curiousity, is there any way (Without l'hopital) to evaluate without a substitution? – 2017-02-28
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    @AspiringMathlete Not really; you can do the same without the substitution, just using $x=(x-\pi/4)+\pi/4$ and at the end you get the same expression, so you end up with $\lim_{x\to\pi/4}\frac{\tan(x-\pi/4)}{x-\pi/4}$. Or as I added; but you always need $\lim_{x\to a}\frac{\sin(x-a)}{x-a}=1$. – 2017-02-28
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$$\lim_{x \to \frac{\pi}{4}} \frac{1-\tan x}{x-\frac{\pi}{4}} =\lim_{x \to \frac{\pi}{4}} \frac{-\tan (x-\frac{\pi}{4})}{x-\frac{\pi}{4}}(1+ \tan x)=-1(2)=-2$$

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    I don't see what you did in the second step... – 2017-02-28
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    @AspiringMathlete Note that $${-\tan (x-\frac{\pi}{4})}=\frac{1 - \tan x}{1+ \tan x}$$ – 2017-02-28