Math, asked by debashissaha401, 4 months ago

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

Attachments:

Answers

Answered by ExploringMathematics
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}}

Answered by KingSrikar
0

WELL, IT'S AN INTEGRATION QUESTIO

Similar questions