...............................1) lim x->INFINITY
arctan(x^6 - x^8)
the answer is -pi/2, but i dont know how to get it
== Arctan is the inverse function of tan and tan x --> 00 if x --> Pi/2, and tan x --> -oo if x --> -Pi/2 (this is basic trigonometry, nothing fancy).
From here what you get.
2) on pic (it will have a red number 2 on it)
== Just apply the def. of derivative to the function f(x) = 2^x at the point x = 4.
3) on second pic (it will have a red number 3 on it)
in case the limt exists, we know f ' (0) = lim [f(x) - f(0)]/x when x --> 0, so:
lim [x sin(8/x) - 0]/x = lim sin(8/x) , and this doesn't exist since 8/x --> oo when x --> 0 from the right, and thus sin(8/x) swings between -1 and 1 without approaching any definite number (for example, choose x_n = 1/nPi so that 8/x_n = 8nPi, n an integer ==> sin(8/x_n) = 0 and clearly x_n --> 0 when n --> oo. Now choose y_n = 1/(n/16)Pi, with n an odd integer. again, y_n --> o when n-->oo, but this time sin(8/y_n) = (n/2)Pi, and since n is odd sin(8/y_n) is 1 or -1 ==> again, the limit doesn't exist.
for 2 and 3, both answers are on it, but i have no idea into how to get this
does anyone know?????????