Is the limit of exponential function y = a ^ x, X positive infinity and X -- negative infinity infinite and 0 respectively?

No, it depends on the range of a, a = 1, and the value of the function is always 1

How to prove that the exponential function of e tends to positive infinity with the definition of limit?

If n = ln (n), then M > N, exp (m) > exp (n) = exp (lnn) = n, so exp (x) tends to positive infinity when x tends to positive infinity

Given the function f (x) = 3 to the power X, and f (a + 2) = 18, G (x) = 3 to the power ax

The definition field of the power ax of G (x) = 3 and the power X of - 4 is [0,1]
(1) The analytic formula of G (x)
(2) Find the monotone interval of G (x), determine its increase and decrease, and prove it by definition

】It is known that the function f (x) = the third power of X + ax + X + B, where a and B belong to R
If f (x) is an increasing function on [0,2], the value range of a is obtained

If f '(x) = 3x ^ 2 + 2aX + 1 is an increasing function on [0,2], then in this interval, f' (x) > = 0, i.e., 3x ^ 2 + 2aX + 1 > = 0 is obviously satisfied. When x > 0, a > = - (3x ^ 2 + 1) / (2x) = g (x), then find the maximum value of G (x), G (x) = - (1.5x + 0.5 / x) from the mean inequality, 1.5x + 0.5 / x > = 2 √ (1.5x * 0.5 / x) = √ 3 when 1.5x = 0.5 / x

Is the limit of exponential function y = a ^ x, X positive infinity and X -- negative infinity infinite and 0 respectively?