lim x→0 ((18x - 2sin 5x)/(tan10x - 3sin4x) )
lim x→¼π ( 8cosx - 8sinx )/(x - π/4)
lim x→¼π ( 8cosx - 8sinx )/(x - π/4)
-
Using L'hôpital's rule
We have limit (f(x) / g(x)) as x tends to c
If f and g tend to 0 and the derivative of them exists then
(f(x) / g(x)) as x tends to c = (f'(x) / g'(x)) as x tends to c
a)
f'(x) = 18-10*cos(5*x)
f'(0) = 18 - 10 = 8
g'(x) = 10+10*tan²(10*x)-12*cos(4*x)
g'(0) = 10 + 0 - 12 = -2
f'(0) / g'(0) = -4.
b)
f'(x) = -8*sin(x)-8*cos(x)
f'(π/4) = -8√2
g'(x) = 1
g'(π/4) = 1
f'(π/4) / g'(π/4) = -8√2
We have limit (f(x) / g(x)) as x tends to c
If f and g tend to 0 and the derivative of them exists then
(f(x) / g(x)) as x tends to c = (f'(x) / g'(x)) as x tends to c
a)
f'(x) = 18-10*cos(5*x)
f'(0) = 18 - 10 = 8
g'(x) = 10+10*tan²(10*x)-12*cos(4*x)
g'(0) = 10 + 0 - 12 = -2
f'(0) / g'(0) = -4.
b)
f'(x) = -8*sin(x)-8*cos(x)
f'(π/4) = -8√2
g'(x) = 1
g'(π/4) = 1
f'(π/4) / g'(π/4) = -8√2