Limit x tends to 0 [tanx/x]^1/x^2 is ?lim m tends to infinity [cos (x/m)] ^m is?I need the procedure of solving these sums.. Thank you

In These sums put yhe value of x like in first question tanx/x when x tens 0 is 1 and i/x2 becomes infinity ie it is of the form 1 raised to the infinity

a simple formula is Lt x tends 0 f(x)^g(x) where it is of the form 1^inf is equal to e^Lt x tends 0 ((f(x) -1)/G(x))

for first it becomes e^ ( lt x tends 0 tanx -x/x^3)

apply L hopital twice it becomes e^1/3sec^2x tanx/x

= e^1/3

Approved

Dear Shuvam,
The question is not in INDETERMINATE FORM.
you need to convert it in that form by using a simple formula...

:-  lim x->a   [ 1 + f(x)]^g(x)  =   e^lim x->a  f(x).g(x)

apply this formula and you get the ans as e^1/2 in the first one.

for the second put m=1/n
hence when m->infinity.... n->0
and then apply the same formula
you get the ans as 1.


Regards.

Approved

Let me give you a brief idea:

It is in 1^infinity form.

If f(x)=g(x)=0 when limit x tends to 0:

lt_{x tends 0}f(x)^g(x)=e^{lt ((f(x)-1)g(x)}

apply this formula and solve by l hospital rule. 

Approved

as x tends to zero tanx/x tends to 1 but 1/x^2 tends to infinity.Hence, it is 1^infinity. so we use a standard result which is (1+(tanx/x)-1)^the reciprocal*1/x^2 . Hence we have the limit is e^1/3

Approved

e^1/3