Snehal Jadhav
Last Activity: 6 Years ago
The question is of the form 1infinity.
Such type of questions can be solved by using formula:
f(x)g(x) = e[f(x)-1]*g(x)
We, will evaluate e raised to the power:
(2 – (x/a) – 1)*tan(

x/2a)
(1 – (x/a))*tan(

x/2a)
((a – x)/a)*tan(

x/2a)
Now, we will change the limits from x tends to a to (a+h) where, h tends to zero.
[(a – (a+h))/a]*tan(

(a+h)/2a)
[ – h/a]*[tan(

/2 +

h/2a)
(-)h/a*(-)*(1/tan(

h/2a))
h/a*(1/tan(

h/2a))
The term
tan(
h/2a) tends to zero as h tends to zero.
So, tan(

h/2a) equals to
(
h/2a).Hence, the terms h and a cancel out.
2/h
So, the answer is: e2/h