1) \lim_{x\rightarrow \Pi /2} (sinx)^{tanx}
2) \lim_{x\rightarrow \propto } (\frac{x^{2}-2x+2}{x^{2}-4x+1})^{x}
231st one is 1∞ form .
so
logL = \lim_{x\rightarrow \pi /2}(tanx)log(sinx)
L=e^{\lim_{x\rightarrow \pi /2} (logsinx)(tanx)}
L=e^{\lim_{x\rightarrow \pi /2}\frac{log(sinx)}{cotx}}
the lim is of 0/0 form . So apply L hospital.
try second one in the same manner
6@ avik i have given a formula to solve 1∞form questions.
check in ur other limit threads.such forms are called power limit.
11) limit doesn't exist..
coz say if you approach x from it's LHS sinx<1 and tanx->∞.. net answer zero..
but if you approach x from RHS sinx is still <1 but tanx is now --> -∞.
so in that case your ans is ∞..
HENCE, LIMIT DOESN'T EXIST..
2)e2
6solution
divide num nd denm by x2
you get limx→∞{(1-1/x+/x2)/1-4/x+1/x2}x
putting x = ∞ we get the form 1∞
so we can use the power limit formula i told u
then it becomes
e{x2-2x+2/x2-4x+1-1}*x
then simplify nd u will get e2
hope this is d ans nd procedure