39given limit = elimx→a[2-x/a -1 ]tan πa/2x [ the limit is in the power of e]
limx→a {1-x/a}tan πa/2x =limx→a tan πx/2a / a/a-x this is of the form ∞/∞ and can be evaluated by l hospital`s rule.
13 Answers
3939i think the limit does not exist as on continuosly using l`hospital`s rule its remaining in ∞/∞ form
whats the answer rohit?
1even i got that.........though i wasnt sure..as it was a skul paper.............dunno d answer,,,,dude,,,thanxn....
1yah..i proved by both ways..that limits doesnt exist..make it of d form
1 + f(x).........
1hint take tanx=sinx/cosx.....then apply l hospitals to (1-x/a)/cosx (0/0 form)....it will be 2/Î ...hence the answer..
341Limx→a [1 + (1- x/a)]tan (πx/2a)
= Lim1-x/a→0 [1 + (1- x/a)]1/(1-x/a)tan (πx/2a) (1-x/a)
= eLim x→a tan (πx/2a) (1-x/a)
Now, Lim x→a tan (πx/2a) (1-x/a)
= Limπ/2 (1-x/a)→0 [π/2 (1-x/a)]/[tan π/2 (1-x/a)] * 2/π
= 2/Ï€
Hence, the given limit is 2/Ï€
1357Rohit......Regarding your doubt in last part
lim(x→a) tan(πx/2a) (1-x/a)
lim(x→a) (1-x/a)/ cot(πx/2a)
lim(x→a) (1-x/a)/ tan(π/2-πx/2a)
let π/2(1-x/a)=t
it becomes
lim(t→0)(2/π)(t/tant)
1kya bhaiya! mere answer ko kabhi kabhi pink kar dijiye...
...:)
arey bhai! confidence badhta hai...:)
