Limit: when x tends to infinity [ln (2 ^ x + 3 ^ x)] / [ln (3 ^ x + 4 ^ x)]

Limit: when x tends to infinity [ln (2 ^ x + 3 ^ x)] / [ln (3 ^ x + 4 ^ x)]

The results were: Ln3 / ln4
Use the lobita rule first
Original form=
lim[( ln2 *2^x +ln3 3^x)/(2^x+3^x)]*[(3^x+4^x)/(ln3 *3^x+ln4 *4^x)]
For the first one, the numerator and denominator are divided by 3 ^ x at the same time. For the second one, the numerator and denominator are divided by 4 ^ x at the same time
Results: Ln3 / ln4

Given f (x) = AX3 + bx-4, if f (- 2) = 2, then f (2) = ()
A. -2B. -4C. -6D. -10

∵ f (x) = AX3 + bx-4, ∵ f (- x) + F (x) = - ax3-bx-4 + AX3 + bx-4 = - 8, ∵ f (- 2) = 2, ∵ 2 + F (2) = - 8, the solution is f (2) = - 10

Given that a and B are constants, and the limit of [(AX + b) / (x + 1)] is equal to 3, X tends to - 1, find a and B

When x tends to - 1, the denominator limit is 0, so - A + B = 0,
Because the limit is 3, ax + B = 3 (x + 1) = 3x + 3,
The solution is a = 3, B = 3