Question

the value of limit x tends to zero cosx/π/2-x is​

Answer 1

$\rm{\displaystyle\lim _{n\to \:0}\left(\frac{\cos \left(n\right)}{\frac{\pi }{2}-x}\right)=\frac{\cos \left(0\right)}{\frac{\pi }{2}-x}\quad...\:Plugging\:in\:the\:value\:n=0}$

$\longrightarrow\rm{\displaystyle\lim _{n\to \:0}\left(\frac{\cos \left(n\right)}{\frac{\pi }{2}-x}\right)=\frac{1}{\frac{\pi }{2}-x}\quad...\:Since\:cos(0)=1}$

$\longrightarrow\rm{\displaystyle\lim _{n\to \:0}\left(\frac{\cos \left(n\right)}{\frac{\pi }{2}-x}\right)=\frac{1}{\frac{\pi }{2}-\frac{2x}{2}}}\longrightarrow\rm{\displaystyle\lim _{n\to \:0}\left(\frac{\cos \left(n\right)}{\frac{\pi }{2}-x}\right)=\frac{1}{\frac{\pi -2x}{2}}}$

$\longrightarrow\rm{\displaystyle\lim _{n\to \:0}\left(\frac{\cos \left(n\right)}{\frac{\pi }{2}-x}\right)=\frac{2}{\pi -2x}}$

Answer 2

WELL, IT'S AN INTEGRATION QUESTIO