calculus problem on limits

Question

Comments

Answer: 1 ?

Answer 1

It is 0^0 formate apply L hospital

= e ^ limt x-> 0 (sinx ln x)

e^ limt x->0 (lnx/cosecx) apply L hospital

e ^ limt x->0 (-(sinx/x cotx))

e ^ 0

=1

Answer 2

= lt_x->0 x ^sinx

Let Y = X^sinx

log ^Y = sinx log ^x ; // apply log on both the sides

lt_{x -> 0} log^Y = lt_x->0  sinx logx ; // apply limit on both sides

log^Y = lt_{x-> 0} logx/cosecx; // infinity / infinity ; so apply L Hospital rule ; lt_{x -> 0} log^Y = log ^Y why because here log ^Y is constant doesn't depend upon x.

log^{Y  =} lt_x->0 (1/x)/-cosecx.cotx ; //

log^Y = lt_x->0(-sinx. tanx)/x ; //

log^Y = lt_x->0 -sinx. (tanx/x) ;

log^Y = lt_x->0 -sinx . 1 ; // lt_x->0 tanx/x = 1

log ^{Y =} 0

Y = e⁰ = 1

therefore, Y = 1

Answer 3

1 fx power gx type function

Answer 4

DIRectly substitute value of x

Answer is 1

Answer 5

This problem is in 0^0 form so, we cant apply L Hopital Rule .

Please follow the below step and you will get the answer.

Take Log on both the side and bring the limit in O/O form .

Apply L Hopital and solve further.