Question Number 215667 by mathlove last updated on 14/Jan/25
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{sec}\left(\frac{\pi}{\mathrm{2}}{x}\right)\left({arctanx}−\frac{\pi}{\mathrm{4}}\right)=? \\ $$
Answered by issac last updated on 14/Jan/25
$$\mathrm{sec}\left(\frac{\pi}{\mathrm{2}}\right)\left(\mathrm{arctan}\left(\mathrm{1}\right)−\frac{\pi}{\mathrm{4}}\right) \\ $$
Commented by Ghisom last updated on 14/Jan/25
$$\mathrm{just}\:\mathrm{tell}\:\mathrm{us}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sec}\:\frac{\pi}{\mathrm{2}} \\ $$
Commented by mathlove last updated on 14/Jan/25
$${corect}\:{answer}\:{is}\:−\frac{\mathrm{1}}{\pi} \\ $$
Answered by Ghisom last updated on 14/Jan/25
![lim_(x→1) (arctan x −(π/4))sec ((πx)/2) = =lim_(x→1) ((arctan x −(π/4))/(cos ((πx)/2))) = [l′Ho^ pital] =lim_(x→1) ((1/(x^2 +1))/(−(π/2)sin ((πx)/2))) =((1/2)/(−(π/2)))=−(1/π)](https://www.tinkutara.com/question/Q215671.png)
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\mathrm{arctan}\:{x}\:−\frac{\pi}{\mathrm{4}}\right)\mathrm{sec}\:\frac{\pi{x}}{\mathrm{2}}\:= \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{arctan}\:{x}\:−\frac{\pi}{\mathrm{4}}}{\mathrm{cos}\:\frac{\pi{x}}{\mathrm{2}}}\:= \\ $$$$\:\:\:\:\:\left[\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}}{−\frac{\pi}{\mathrm{2}}\mathrm{sin}\:\frac{\pi{x}}{\mathrm{2}}}\:=\frac{\frac{\mathrm{1}}{\mathrm{2}}}{−\frac{\pi}{\mathrm{2}}}=−\frac{\mathrm{1}}{\pi} \\ $$