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Can someone help to do the following limit without L'Hospital or series expansion:

$$\lim_{x\to 2} \frac{\arctan x^2-\arctan 4}{\tan 2^x-\tan 4}$$

With l'Hospital it is simple:

$$\lim_{x\to 2} \frac{\arctan x^2-\arctan 4}{\tan 2^x-\tan 4}=\lim_{x\to 2} \frac{\frac{2x}{x^4+1}}{2^x \log 2\sec^2(2^x)}=\frac{1}{17\log 2\sec^2(4)}$$

but I have to solve it without l'Hospital and without series expansion. I tried to solve it with $\lim\limits_{x\to 0} \frac{\tan x}{x}=1$, but I can't do it.

LHF
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2 Answers2

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Divide top and bottom by $x-2$ and it becomes a ratio of two limits that are both by definition a derivative,$$\lim_{x\to2}\frac{\frac{\arctan x^2-\arctan4}{x-2}}{\frac{\tan2^x-\tan4}{x-2}}=\frac{\lim_{x\to2}\frac{\arctan x^2-\arctan4}{x-2}}{\lim_{x\to2}\frac{\tan2^x-\tan4}{x-2}}=\frac{\left.\frac{d}{dx}\arctan x^2\right|_{x=2}}{\left.\frac{d}{dx}\tan2^x\right|_{x=2}}.$$The rest is your existing calculation, but this does not use L'Hôpital's rule.

J.G.
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  • It feels like cheating to use derivatives when the question asks without l'Hospital :P (I'm just joking). – LHF Feb 24 '20 at 18:58
  • @Atticus See, this is why I always hate these questions. Unless you can solve it with a clever, say, trigonometric identity, it boils down to a legal argument about whether using the rule's proof strategy, with a special case for the problem, is (a) breaking the spirit of the rule's prohibition or (b) unobjectionable because you're using the first-principles definition of a derivative. I lean toward option (b). – J.G. Feb 24 '20 at 19:03
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    I also agree that your solution is legit in the given restrictions :) – LHF Feb 24 '20 at 19:04
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We can use $\lim\limits_{x\to 0} \dfrac{\tan x}{x}= 1$ and

$$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}$$

to get rid of the $\arctan$s:

$$\begin{aligned} \lim_{x\to 2} \frac{\arctan x^2-\arctan 4}{\tan 2^x-\tan 4} &= \lim_{x\to 2} \left[\frac{\arctan x^2-\arctan 4}{\tan(\arctan x^2-\arctan 4)}\cdot \frac{\tan(\arctan x^2-\arctan 4)}{\tan 2^x-\tan 4}\right]\\ &= \lim_{x\to 2}\frac{\tan(\arctan x^2-\arctan 4)}{\tan 2^x-\tan 4}\\ &= \lim_{x\to 2} \frac{\frac{x^2-4}{1+4x^2}}{\frac{\sin2^x\cos 4-\sin 4\cos2^x}{\cos^{2^x}\cos 4}}\\ &=\frac{\cos^2 4}{17} \cdot \lim_{x\to 2}\frac{x^2-4}{\sin(2^x-4)} \end{aligned} $$

and:

$$ \begin{aligned} \lim_{x\to 2}\frac{x^2-4}{\sin(2^x-4)} &= \lim_{x\to 2} \frac{x^2-4}{2^x-4} \cdot \lim_{x\to 2} \frac{2^x-4}{\sin(2^x-4)}\\ &=\lim_{x\to 2} \frac{x^2-4}{2^x-4}\\ &= \lim_{x\to 2} \frac{x+2}{4} \cdot \lim_{x\to 2} \frac{x-2}{2^{x-2}-1}\\ &=1\cdot \frac{1}{\ln 2} \end{aligned} $$

At the end, I used $\lim\limits_{x\to 0} \dfrac{a^x-1}{x}=\ln a$. Thus, the final answer is $\dfrac{\cos^2 4}{17 \ln 2}$.

LHF
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