 Hello and how are you all doing today? The question says evaluate the following limits. Here the limit given to us is limit x approaches pi by 2 tan 2x divided by x minus pi by 2. And before proceeding on we should be well versed that when limit x approaches 0 then the value of sin x over x is equal to 1 and this is the key idea for our question. Now proceeding on with the solution we have limit x approaches pi by 2 tan 2x divided by x minus pi by 2. Now here let us put x minus pi by 2 equal to h. So as x approaches pi by 2 therefore h will approach to 0, right? So now we have limit h approaches 0 tan 2 now in place of x we will substitute h plus pi by 2 divided by x minus pi by 2 is substituted as h. Further we have limit h approaches 0 tan 2h plus pi divided by h. Further now we can evaluate it as limit h approaches 0. Now tan 2h plus pi is what tan 2h itself so we have tan 2h divided by h. Now also we know that tan 2h is what it is sin 2h divided by cos 2h right? The whole divided by h which can be splitted and written as limit h approaches 0 sin 2h divided by h multiplied by 1 over cos 2h. Further we have limit h approaches 0. Now this can be written as sin 2h divided by 2h multiplied by 1 plus 1 divided by cos 2h. Now since h approaches to 0 2h will also approach to 0 so we have limit 2h approaches 0 the constant will come out sin 2h upon 2h multiplied by limit h approaches 0 1 over cos 2h. Now here as we know that limit x approaches 0 the value of sin x over x is equal to 1. On using this key idea we have over here 2 into 1 into 1 over 1 that is further equal to 2. So the answer to the session is 2 that is the value of the limit when x approaches pi by 2 tan 2x upon x minus pi by 2. This ends my session hope you understood the concept well have a nice day ahead.