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  • Limit with tan and <img alt="\mathrm{\arctan \lim _{x \rightarrow 2} \frac{\arctan x^{2}-\arctan 4}{\tan 2^{x}-\tan 4}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Carctan%2

Limit with tan and \mathrm{\arctan \lim _{x \rightarrow 2} \frac{\arctan x^{2}-\arctan 4}{\tan 2^{x}-\tan 4}}

Option: 1

\frac{\cos ^{2} 4}{17 \ln 2}


Option: 2

0


Option: 3

\cos ^{2} 4


Option: 4

None


Answers (1)

best_answer

We can use \mathrm{\lim _{x \rightarrow 0} \frac{\tan x}{x}=1} and
\mathrm{\tan (a-b)=\frac{\tan a-\tan b}{1+\tan a \tan b}}

to get rid of the arctans:
\mathrm{\lim _{x \rightarrow 2} \frac{\arctan x^{2}-\arctan 4}{\tan 2^{x}-\tan 4} =\lim _{x \rightarrow 2}\left[\frac{\arctan x^{2}-\arctan 4}{\tan \left(\arctan x^{2}-\arctan 4\right)} \cdot \frac{\tan \left(\arctan x^{2}-\arctan 4\right)}{\tan 2^{x}-\tan 4}\right]}
                                                      \mathrm{=\lim _{x \rightarrow 2} \frac{\tan \left(\arctan x^{2}-\arctan 4\right)}{\tan 2^{x}-\tan 4}}
                                                      \mathrm{=\lim _{x \rightarrow 2} \frac{\frac{x^{2}-4}{1+4 x^{2}}}{\frac{\sin 2^{2} \cos 4-\sin 4 \cos 2^{x}}{\cos ^{2} \cos 4}}}
                                                      \mathrm{=\frac{\cos ^{2} 4}{17} \cdot \lim _{x \rightarrow 2} \frac{x^{2}-4}{\sin \left(2^{x}-4\right)}}
and:
\mathrm{\lim _{x \rightarrow 2} \frac{x^{2}-4}{\sin \left(2^{x}-4\right)} =\lim _{x \rightarrow 2} \frac{x^{2}-4}{2^{x}-4} \cdot \lim _{x \rightarrow 2} \frac{2^{x}-4}{\sin \left(2^{x}-4\right)} }
\mathrm{=\lim _{x \rightarrow 2} \frac{x^{2}-4}{2^{x}-4} }
\mathrm{ =\lim _{x \rightarrow 2} \frac{x+2}{4} \cdot \lim _{x \rightarrow 2} \frac{x-2}{2^{x-2}-1} }
\mathrm{ =1 \cdot \frac{1}{\ln 2} }

At the end, I used \mathrm{ \lim _{x \rightarrow 0} \frac{a^{x}-1}{x}=\ln a }. Thus, the final answer is \mathrm{ \frac{\cos ^{2} 4}{17 \ln 2} }.

Posted by

vishal kumar

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