Question

I have attached a image with three questions of limit. Kindly answer these questions!

Answer 1

Step-by-step explanation:

Let f (x) = x^(1/x) = e^(ln (x^(1/x))) = e^((ln x)/x),

and then f ’(x) = (x^(1/x)) · (1 – ln x)/(x^2).

f is continuous and differentiable on (0, ∞) .

Now f ’(x) can only equal 0 when

1 – ln x = 0 so x = e is the only critical value.

Furthermore, f ’(x) > 0 on (0, e) and so f increases on (0, e)

and f ’(x) < 0 on (e, ∞) and so f decreases on (e, ∞);

therefore f has an absolute maximum at x = e.

Thus, for integer values of x, either f (2) or f (3) must be the maximum of the given series.

Now if both 2^(1/2) and 3^(1/3) were raised to the sixth power we would have

(2^(1/2))^6 = 2^3 = 8 which is less than (3^(1/3))^6 = 3^2 =9.

Since y = x^6 is monotonically increasing on (0, ∞) and 8 < 9 then 2^(1/2) < 3^(1/3); it follows that 3^(1/3) is the greatest value in the series.

hope it will help you mate

have a great day mate

Answer 2

Answer:

I have to try it...

then give you answers ASPS....