IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

v_gurucharan

1^infinity and 0 * infinity are indeterminate forms...........

how? Huh

prove it...........

venkat_tatolu

Take log(1^infinity)=infinity log(1)

=infinity log(1)=infinity*0 =0/0

Now remove log , we get

1^infinity= e^(0/0) which is indeterminate form

Similarly 2nd one also can be proved

johri_anshuman

there is no point of doubt that 1^infinity=1 but in case of indeterminate form the value is (TENDING TO 1)^infinity

consider the example

lim x--->0 (x+1)^1/x

as x tends to 0 ,   x+1 tends to 1 but does not attain the value 1 and the form is of 1^infinity

when x-->0-    (x+1)--->1 but its value is still less than 1
when x-->0+   (x+1)--->1 but its value is more than 1

when with such a minor difference a number is raised to a power tending to infinity the changes become so much so that they cannot be neglected and hence the limit may or may not be 1

So 1^infinite is considered as indeterminate form.

For 0*infinity consider the example lim x-->infinity (1/x+1)*x

as x tends to infinity 1/(x+1) tends to 0 and does not become zero
as this small number is multiplied with infinity the limit comes to be finite.
so 0* infinity is indeterminate.

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