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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
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->0and then apply the same formulayou get the ans as 1.Regards.
Let me give you a brief idea:
It is in 1infinity form.
If f(x)=g(x)=0 when limit x tends to 0:
ltx tends 0f(x)g(x)=elt ((f(x)-1)g(x)
apply this formula and solve by l hospital rule.
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
e^1/3
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