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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6Set $1=\tan\pi/4$ or set $x-\pi/4=u$ – lab bhattacharjee Feb 28 '17 at 13:41
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2Your expression is $-\frac{\tan x-\tan(\pi/4)}{x-\pi/4}$. The limit is the derivative definition $-\tan'(\pi/4)$. – A.Γ. Feb 28 '17 at 13:45
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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 – Aspiring Mathlete Feb 28 '17 at 13:51
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1@AspiringMathlete Checkout my answer. – The Cryptic Cat Feb 28 '17 at 13:58
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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 – lab bhattacharjee Feb 28 '17 at 16:23
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3Does this answer your question? Limit of $\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} $ without l'hopital's – Jul 30 '21 at 06:41
4 Answers
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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 $-\tan'{\pi\over 4}=-2$
marwalix
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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? – Aspiring Mathlete Feb 28 '17 at 14:03
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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. – Did Feb 28 '17 at 14:17
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@AspiringMathlete: S/he didn't, he just noted that the $1$ which was already there equals $\tan\frac\pi4$. – Vincenzo Oliva Feb 28 '17 at 14:17
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$\tan{\pi\over 4}=1$ does not depend on $x$ so there's no limit in there. – marwalix Feb 28 '17 at 15:42
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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 $$ \lim_{x\to\pi/4}\frac{\sin\left(x-\frac{\pi}{4}\right)}{x-\frac{\pi}{4}}\frac{-\sqrt{2}}{\cos x} $$
egreg
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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? – Aspiring Mathlete Feb 28 '17 at 14:05
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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 $\lim_{x\to a}\frac{\sin(x-a)}{x-a}=1$. – egreg Feb 28 '17 at 14:08
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$$\lim_{x \to \frac{\pi}{4}} \frac{1-\tan x}{x-\frac{\pi}{4}} =\lim_{x \to \frac{\pi}{4}} \frac{-\tan (x-\frac{\pi}{4})}{x-\frac{\pi}{4}}(1+ \tan x)=-1(2)=-2$$
The Cryptic Cat
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1@AspiringMathlete Note that $${-\tan (x-\frac{\pi}{4})}=\frac{1 - \tan x}{1+ \tan x}$$ – The Cryptic Cat Feb 28 '17 at 14:07
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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
$$\tan(x)=\tan (a)+ \left(\tan ^2(a)+1\right)(x-a)+O\left((x-a)^2\right)$$
Claude Leibovici
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1@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 :-) – Claude Leibovici Feb 28 '17 at 13:49
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@ClaudeLeibovici I am also a precalculus student and I hope to learn taylor series in two years. – Jul 30 '21 at 07:14
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1@huzaifaabedeen. I hope that you will enjoy them ! It is a fantastci tool. Good luck in your studies. Cheers :-) – Claude Leibovici Jul 30 '21 at 07:19