Saved Bookmarks
| 1. |
Eind the limitう。itana.ス |
|
Answer» so we can useL'Hospital's rule. d/dx(numerator ) = sec ^2 x -1 = tan ^2 x d/dx(denominator = (x^2 sec ^2 x+ 2x tan x) so tan ^2 x/((x^2 sec ^2 x+ 2x tan x) = (tan x/x) (tan x/( x sec^2 x+ 2 tan x) tan x/x -> 1 and tan x/( x sec ^2 x + 2 tan x) is reciprocal of ( x sec ^2 x + 2 tan x)/ tan x = ( x (1/cos^2x tan x) + 2 = x/(cos x sin x) + 2 = (x/sin x)(1/cos x) + 2 = 1 .1 + 2 = 3 so tan x/( x sec ^2 x + 2 tan x) = 1/3 so tan ^2 x/((x^2 sec ^2 x+ 2x tan x) or given expression = 1/3 |
|